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Tamilnadu Samacheer Kalvi 10th Social Science Solutions History Chapter 3 World War II

Samacheer Kalvi 10th Social Science World War II Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
When did the Japanese formally sign of their surrender?
(a) 2 September, 1945
(b) 2 October, 1945
(c) 15 August, 1945
(d) 12 October, 1945
Answer:
(a) 2 September, 1945

Question 2.
Who initiated the formation of League of Nations?
(a) Roosevelt
(b) Chamberlain
(c) Woodrow Wilson
(d) Baldwin
Answer:
(a) Roosevelt

Question 3.
Where was the Japanese Navy defeated by the US Navy?
(a) Battle of Guadalcanal
(b) Battle of Midway
(c) Battle of Leningrad
(d) Battle of El Alamein
Answer:
(b) Battle of Midway

Question 4.
Where did the US drop its first atomic bomb?
(a) Kavashaki
(b) Innoshima
(c) Hiroshima
(d) Nagasaki
Answer:
(c) Hiroshima

Question 5.
Who were mainly persecuted by Hitler?
(a) Russians
(b) Arabs
(c) Turks
(d) Jews
Answer:
(d) Jews

Question 6.
Which Prime Minister of England who signed the Munich Pact with Germany?
(a) Chamberlain
(b) Winston Churchill
(c) Lloyd George
(d) Stanley Baldwin
Answer:
(a) Chamberlain

Question 7.
When was the Charter of the UN signed?
(a) June 26, 1942
(b) June 26, 1945
(c) January 1, 1942
(d) January 1, 1945
Answer:
(b) June 26, 1945

Question 8.
Where is the headquarters of the International Court of Justice located?
(a) New York
(b) Chicago
(c) London
(d) The Hague
Answer:
(d) The Hague

II. Fill in the blanks

1. Hitler attacked ……………… which was a demilitarized zone.
2. The alliance between Italy, Germany and Japan is known as ………………
3. ……………… started the Lend-Lease programme.
4. Britain Prime Minister ……………… resigned in 1940.
5. Saluting the bravery of the ……………… Churchill said that “Never was so much owed by so many to so few”.
6. ……………… is a device used to find out the enemy aircraft from a distance.
7. The Universal Declaration of Human Rights set forth fundamental human rights in ……………… articles.
8. After the World War II ……………… was voted into power in Great Britain.

Answers:

1. Rhineland
2. Rome – Berlin
3. Roosevelt
4. Chamberlain
5. Royal Air force
6. Radar
7. 30
8. Labour party

III. Choose the correct statement

Question 1.
(i) Banking was a major business activity among Jews.
(ii) Hitler persecuted the Jews.
(iii) In the concentration camps Jews were killed.
(iv) The United Nations has currently 129 member countries in it.
(a) (i) and (ii) are correct
(b) (i) and (Hi) are correct
(c) (iii) and (iv) are correct
(d) (i) is correct and (ii), (iii) and (iv) are wrong
Answer:
(a) (i) and (ii) are correct

Question 2.
Assertion (A): President Roosevelt realised that the United States had to change its policy of isolation.
Reason (R): He started a programme of Lend Lease in 1941.
(a) Both A and R are correct
(b) A is right but R is not the correct reason
(c) Both A and R are wrong
(d) R is right but it has no relevance to A
Answer:
(a) Both A and R are correct

IV. Match the Following

Answers:
A. (v)
B. (iv)
C. (i)
D. (ii)
E. (iii)

V. Answer the questions briefly

Question 1.
Mention the important clauses of the Treaty of Versailles relating to Germany.
Answer:

1. Germany was forced to give up territories to the west, north and east of the German border.
2. Germany had to be disarmed and was allowed to retain only a very restricted army, navy and air force.
3. Germany was expected to pay huge military and civilian cost of the war to the allied nations (approx. $ 25 billion).

Question 2.
Who were the three prominent dictators of the post World War I?
Answer:
The three prominent dictators of the post-World War I were Mussolini (Italy), Hitler (Germany) and Franco (Spain).

Question 3.
How did Hitler get the support from the people of Germany?
Answer:
Hitler was able to sway away the emotions of the German people by his great speeches. He promised them that he will return back the glorious Germany. His racial superiority of the Germans as a pure Aryan race and a deep-rooted hatred for jews made him get the support of his people.

Question 4.
Describe the Pearl Harbour incident.
Answer:
Pearl Harbour incident took place in December 1941 when japan attacked American naval installations in Pearl Harbour, Hawaii, without warning to cripple America’s Pacific fleet. Many battle ships and numerous fighter planes were destroyed. The US declared war on Japan, with Britain and China. This brought together both the Asia Pacific and the European war into one common cause. Most importantly, it brought the United States with its enormous resources into the war as a part of the Allies.

Question 5.
What do you know of Beveridge Report?
Answer:
The Report that was published in the United Kingdom in 1942 to improve the general welfare of the people is called as Beveridge Report. It proposed that the government should provide citizens with adequate income, healthcare, education housing and employment to overcome poverty and disease thereby improve general welfare.

Question 6.
Name the Bretton Woods Twins.
Answer:
The World Bank and the International Monetary Fund.

Question 7.
What are the objectives of the IMF?
Answer:

1. To foster global monetary co-operation
2. To secure Financial Stability
3. To facilitate International Trade
4. To promote high employment and sustainable economic growth.
5. To reduce poverty around the world.

VI. Answer the questions given under each caption

Question 1.
Battle of Stalingrad

(a) When did Germany attack Stalingrad?
Answer:
In August 1942, Germany attacked Stalingrad.

(b) What were the main manufactures of Stalingrad?
Answer:
The main manufactures of Stalingrad were armaments and tractors.

(c) What was the name of the plan formulated by Hitler to attack Stalingrad?
Answer:
Fall Blau or Operation Blue

(d) What is the significance of the Battle of Stalingrad?
Answer:
The people of Russia were grateful for Stalin’s conduct of the war. They regarded him as ‘a prodigy of patience, tenacity and vigilance, almost omnipresent, almost omniscient.

Question 2.
Japanese Aggression In South-east Asia

(a) Name the South-east Asian countries which fell to the Japanese.
Answer:
Guam, the Philippines, Hong Kong, Singapore, Malaya, the Dutch East Indies and the Burma all fell to the Japanese.

(b) Account for the setback of Allies in the Pacific region?
Answer:
The Allies had a setback in the Pacific region because of their inadequate preparation. The local people had to face the atrocities of the Japanese.

(c) What is the significance of Battle of Midway?
Answer:
The U.S. navy defeated the Japanese navy in the Battle of Midway. Thus, the battle is in favour of the Allies.

(d) What happened to the Indians living in Burma?
Answer:
The Indians living in Burma walked all the way to the Indian border facing many hardships. Many died of disease and exhaustion.

Question 3.
General Assembly and Security Council

(a) List the permanent member countries of the Security Council.
Answer:
The United States, Britain, France, Russia and China.

(b) What is the Holocaust?
Answer:
The word ‘holocaust’ is used to describe the genocide of nearly six million Jews by the Germans during the Second World War.

(c) Who was the Chairperson of the UN Commission on Human Rights?
Answer:
The widow of US President Franklin Roosevelt was the chairperson of the UN Commission on Human Rights.

(d) What is meant by veto?
Answer:
A veto is the power to unilaterally stop an official action, especially the enactment of legislation.

VII. Answer in detail

Question 1.
Attempt an essay on the rise and fall of Adolf Hitler.
Answer:

1. Adolf Hitler Was the founder of the National Socialist party, generally known as the Nazi party.
2. His great oratorical skill, his promise to bring back the glorious past of Germany, his support for the German race and hatred towards the Jews helped him to get people support.
3. He came to power in 1933 and ruled Germany till 1945.
4. He began to re-arm Germany and recruitment of new armed forces.
5. The manufacture of armaments and machinery for the army, navy and air force with large spending from government resulted in the revival of the economic condition and helped to solve the unemployment problem in the economy.
6. He followed aggressive policy and therefore in 1936, he invaded Rhine land, the demilitarized zone.
7. His alliance with Italy and Japan became Rome-Berlin-Tokyo axis.
8. He signed Munich pact stating Germany would not conquer any other territory, rather in 1939, he invaded Czechoslovakia.
9. His attack on Poland resulted in the declaration of war by Britain and France against Germany.
10. In 1941, German army invaded Russia. But the resistance of the German army and Russian winter defeated German army.
11. When the allied forces fought back, Germany also retaliated. Finally, Hitler committed suicide in 1945.
12. In 1945, allies occupied Berlin and Germany was divided as two sections after the war.

Question 2.
Analyse the effects of World War II.
Answer:
World War II was the most devastating war in history. It left a deep impact on the entire world. It changed the world in fundamental ways. Here are the effects of this War:

(i) The world got polarised into two main blocs led by superpowers, one led by the United States which followed anti-communist ideology, and the other by Soviet Russia which was essentially communist in nature. Europe was thus divided into two: Communist and non-communist.

(ii) The United States and the Soviet Union entered into a race to have more nuclear powered World War II 43 weapons. They built a large stockpile of such weapons. Meanwhile, Britain and France developed their own nuclear weapons.

(iii) Gradually there arose competition among countries. They began to devote large amount of resources in developing more and more powerful weapons with great destructive power, and defence spending skyrocketed in many countries.

(iv) It was realised that the League of Nations was ineffective and weak. So countries of the world decided not to repeat the mistake. Instead, many international agencies, in particular the United Nations, the World Bank and the International Monetary Fund came into existence providing a forum for countries large and small.

(v) Many other important social and economic changes also took place in the post-War world. Colonial powers were forced to give independence to former colonies in a process of decolonisation. India was the first country to get independence.

(vi) Women became the part of labour force in huge numbers. They became economically independent.

Question 3.
Assess the structure and activities of the UN.
Answer:
The charter of the United Nations was signed on June 26, 1945 by 51 nations. Now, the United Nations has 193 member states and each one has an equal vote in the UN.

Structure of UN:

The General Assembly: Meets once in a year. Issues of interest and points of conflict are discussed in the Assembly.

The Security Council: Consist of five permanent members (USA, Britain, France, Russia, China) and ten non-permanent members (elected in rotation). Each permanent member has the right to veto (A right to reject a decision).

UN Secretariat: Headed by the Secretary by law General. He is elected by the General Assembly on the recommendations of the Security Council. He, with his cabinet and officials run the UN.

International Court of Justice: Headquarters at the Hague in Holland.

The Economic and the Social Council: Co-ordinates all the social and economic work of the U.N. Headed by economists like Gunnar Myrdal.

Activities of the UN:

1. Human Rights, Refugees problem, climatic change, gender equality are the important issues taken over and deals with it. Earlier in 1960’s decolonisation was also a part of their activity.
2. UN peace keeping force acted in many areas of conflict all over the World. Indian army has been a part of it.
3. The preamble of the UN declares, its activities include human rights, equality of men and women.

VIII. Students Activity

Question 1.
A debate in the class on the success or failure of the UN in preserving World Peace.
Answer:
The students can take the following topics for debate and finally conclude, UN is successful as it has stopped the nations from bringing another war. Small to big clashes were/are handled by UN efficiently.

Argument for:
The topics of discussion for debate are:

1. Solving International conflicts: Since 1945, UN peacekeepers have undertaken over 60 field missions and negotiated 172 peaceful settlements that ended Regional conflicts.
2. Liberation from Colonial rule: Eighty nations and more than 750 million people have been freed from colonialism.
3. Human Rights: Custodian for the protection of human rights, discrimination against women, Children’s rights, torture, missing persons etc. in many countries.
4. Enhancing Human life: Specialised agencies of the UN engaged in enhancing all aspects of human life, including education, health, poverty reduction, climate change.
5. Treaties: More than 560 multilateral treaties on human rights, refugees, disarmament.

Argument against:

Non-proliferation Treaty (NPT): Signed by 190 nations, all live superpowers owned nuclear weapons. Later, several countries North Korea, Israel, Pakistan, India developed nuclear weapons.

Veto Power: Veto power has limited its effectiveness at critical times.

War Criminals: The International criminal court has prosecuted several war criminals. But it has been criticised for prosecuting only African leaders. But Western powers too have committed war crimes.

Israel Attack: Israel attacked homes schools, U.N. shelters in Gaza killing 2,200 Palestinians. The U.N. Security Council has failed any action against Israel.

Conclusion: U.N. is imperfect but it is also indispensable. It is successful as, it is avoiding any other war.

Question 3.
Marking the Allies and Axis countries, as well as important battlefields of World War II in a world map.
Answer:

IX. Map Work

Question 1.
Mark the following on the world map.
1. Axis Power Countries
2. Allied Power Countries
3. Hiroshima, Nagasaki, Hawai Island, Moscow, San Fransico
Answer:
(2)

(3)

Timeline:

Samacheer Kalvi 10th Social Science World War II Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The financial cost of the II World War was ……………….. times higher than that of the I World War.
(a) one
(b) three
(c) five
(d) seven
Answer:
(c) five

Question 2.
The coal mines given to France were ………
(a) Jharia
(b) Saar
(c) Bokaro
Answer:
(b) Saar

Question 3.
“Money in wheelbarrows to buy bread” in the 1920’s. Which country referred to here.
(a) Italy
(b) Austria
(c) Germany
(d) Spain
Answer:
(c) Germany

Question 4.
The principles of war and conquests was glorified by ………
(a) Moderates
(b) Dictators
(c) Extremists
Answer:
(b) Dictators

Question 5.
Hitler broke the Munich pact by invading ……………….. in 1939.
(a) Manchuria
(b) Sudetenland
(c) Poland
(d) Czechoslovakia
Answer:
(d) Czechoslovakia

Question 6.
Hitler demanded the surrender of ………
(a) Danzig
(b) Jutland
(c) Estonia
Answer:
(a) Danzig

Question 7.
The attack of ……………….. by Germany was the final act which result in the initiation of II World War.
(a) Britain
(b) France
(c) Russia
(d) Poland
Answer:
(d) Poland

Question 8.
The British Prime Minister during the Second World War was ………
(a) Sir Winston Churchill
(b) Clement Atlee
(c) Lloyd George
Answer:
(a) Sir Winston Churchill

Question 9.
The tactic followed by Germany to overrun other countries was called as:
(a) Sea-borne invasion
(b) Blitzkrieg
(c) Dunkirk
(d) None
Answer:
(b) Blitzkrieg

Question 10.
In ………, Hitler invaded Russia.
(a) 1940
(b) 1941
(c) 1943
Answer:
(b) 1941

Question 11.
“We shall fight in the fields and in the streets” ……………….. but, we shall never surrender.”- said by
(a) Winston Churchill
(b) Napoleon Bonaparte
(c) George Washington
(d) Roosevelt.
Answer:
(a) Winston Churchill

Question 12.
……………….. used the device radar for detecting aircraft at a distance in World War II.
(a) Germany
(b) Japan
(c) Britain
(d) USA
Answer:
(c) Britain

Question 13.
In September 1940, London was bombed mercilessly by German Air force. This action was called as:
(a) Spit fires
(b) Hurricanes
(c) Blitz
(d) Dunkirk
Answer:
(c) Blitz

Question 14.
Land lease programme of USA took place between the years:
(a) 1939 – 1945
(b) 1941 – 1945
(c) 1936 – 1940
(d) 1914 – 1918
Answer:
(b) 1941 – 1945

Question 15.
In the war between Germany and Russia in 1941, ……………….. was defeated.
(a) Germany
(b) Russia
(c) Britain
(d) None
Answer:
(a) Germany

Question 16.
Stalingrad I situated along the banks of the river:
(a) Miami
(b) Volga
(c) Hwang-Ho
(d) Marne
Answer:
(b) Volga

Question 17.
In the battle of Stalingrad, Germans used the code word ……………….. on Russia.
(a) Alamein
(b) Land lease
(c) Fall Blau
(d) Montegomary
Answer:
(c) Fall Blau

Question 18.
Mussolini was killed by a ……………….. partisan.
(a) Germany
(b) Italy
(c) Russia
(d) Britain
Answer:
(b) Italy

Question 19.
Mussolini was killed in:
(a) May 1945
(b) April 1944
(c) April 1945
(d) May 1946
Answer:
(c) April 1945

Question 20.
In 1945 ……………….. was divided into two sections.
(a) Germany
(b) Italy
(c) Bengal
(d) Russia
Answer:
(a) Germany

Question 21.
Japanese army indulged in the biggest slaughter in the place ……………….. in China.
(a) Manchuria
(b) Nanking
(c) Peking
(d) Shangai
Answer:
(b) Nanking

Question 22.
Japan announced surrendered to U.S on ……………….. 1945.
(a) 2nd September
(b) 15th August
(c) 3rd August
(d) 5th February
Answer:
(b) 15th August

Question 23.
The Security council has ……………….. members.
(a) 10
(b) 15
(c) 25
(d) 3
Answer:
(b) 15

Question 24.
At present, the United Nations has ……………….. member states.
(a) 196
(b) 195
(c) 194
(d) 193
Answer:
(d) 193

Question 25.
The World Bank is located at:
(a) Sweden
(b) New Zealand
(c) Washington
(d) New York
Answer:
(c) Washington

Question 26.
IMF has at present ……………….. member countries.
(a) 200
(b) 187
(c) 189
(d) 190
Answer:
(c) 189

Question 27.
IMF help the countries to solve their ……………….. position.
(a) debt
(b) Balance of payment
(c) Independency
(d) Trade
Answer:
(b) Balance of payment

Question 28.
The report published in 1942, in United Kingdom for the general welfare of the people was called as ……………….. report.
(a) Bretton Woods
(b) Beveridge
(c) Blitzkrieg
(d) Common wealth
Answer:
(b) Beveridge

Question 29.
……………….. party in Great Britain promised for a welfare state to the people.
(a) Communist party
(b) Democratic party
(c) Socialised party
(d) Labour party
Answer:
(d) Labour party

Question 30.
The benefits to the people can be achieved either through ……………….. transfers or free services.
(a) Cash
(b) Country
(c) State
(d) Regional
Answer:
(a) Cash

II. Fill in the blanks

1. World War II began in ……………… and ended in ………………
2. The Treaty of ……………… was signed at the end of World War I in 1919.
3. The Germans offered to pay ……………… billion gold marks to the allies.
4. The United States was faced with great depression after ………………
5. The National Socialist party in Germany was generally known as ………………
6. In 1938, Hitler signed the Munich pact with Prime Minister ………………
7. In 1939, Hitler invaded ……………… as against his promise in Munich pact.
8. Hitler showed hatred against ………………
9. Hitler came to power in ……………… and ruled till ………………
10. World War II was a ……………… war fought with tanks, submarines, bomber planes etc.
11. Britain and France declared war on Germany in ………………
12. In ………………, Italy and Japan joined the axis powers.
13. In September 1940, London war bombed by Germans mercilessly. This action was known as ………………
14. Blitzkrieg means ………………
15. The name of the Britain navy was ………………
16. The war between Britain and ……………… took place in Dunkirk in 1940.
17. The fighter planes of the British Royal force was called as ……………… and ………………
18. ……………… of America started the Land Lease programme.
19. Caucasus was famous for its ……………… in Russia.
20. Mussolini of Italy was killed by an ……………… partisan.
21. The battle of ……………… was considered to be the Great patriotic war by the Russians.
22. Italy surrendered to the allies in ………………
23. The Allied forces under the command of ……………… invaded Normanday in France.
24. Canton was called as ……………… in China.
25. On December 1941, ……………… attacked American naval installations in Pearl Harbour.
26. Guadalcanal is in the ……………… islands.
27. USA dropped an atomic bomb on ……………… and ……………… cities of Japan.
28. Japan announced their surrender on ………………
29. Japan formally signed their surrender marking the end of the World War II was ………………
30. ……………… and ……………… are the two super powers after the II World War.
31. US and Soviet Russia entered into a race to have more ………………
32. ………………, ………………, and ……………… came into existence after the II World War.
33. ……………… started entering into labour force in huge number after World War II.
34. In the process ofdecolonisation ……………… was the first country toget Independence.
35. The word ……………… refers the genocide of Jews by the Germans during Second World War.
36. A major outcome of the Holocaust was the creation of the State of ………………
37. ……………… became the Homeland for Jews after II World War.
38. The Un efforts to protect human rights at the global levei resulted in the UN commission on ………………
39. The Un adopted the Human Rights Charter on ………………
40. ……………… is observed globally as Human Rights Day.
41. Britain and United States gave a joint declaration called as ……………… in 1941 that helped in the formation of UNO.
42. ……………… were the axis powers of the II World War.
43. The initial member States of the UN were ……………… nations.
44. The Charter of the United Nations was signed on ………………
45. Each member State in U.N.has ……………… vote.
46. The UN functions almost like a ………………
47. There are ………………, ………………, ………………, ……………… wing for the UN.
48. Veto means ………………
49. ……………… has veto power.
50. ……………… permanent members are there in UN.
51. WHO means ………………
52. UNICEF means ………………
53. FAO means ………………
54. UNESCO expansion is ………………
55. UNDP expansion is ………………
56. The ……………… has been a port of peace keeping force of the UN in deployment to many parts of the World.
57. The World Bank and the IMP are referred to as ………………
58. The two main organs of the World Bank are ……………… and ………………
59. IBRD expansion is ………………
60. IDA expansion is ………………
61. The IDA lends money to the ……………… for development activities.
62. The loans sanctioned by IDA at low interest rates for development purposes are called as ………………
63. Soft loans are given for ……………… years.
64. The ……………… functions with private enterprises in developing countries.
65. IFC expansion is ………………
66. The World Bank is actively promoting the cause of improving the and eradicating the ………………
67. The IMF was the brainchild of ……………… and ………………
68. The initial member countries of IMF were ………………
69. Its primary objective is to ensure ……………… and development across the World.
70. The fund gives resources to countries facing ……………… problem.
71. The number of member countries of IMF at present are ……………… countries.
72. All the countries in the Western Europe are now ………………
73. The ……………… in Great Britain after World War I promised to look at the people from the cradle to the grave.
74. Legislations was enacted to provide comprehensive free health coverage to the citizens in Britain through ………………
75. The monetary benefits after World War II by Labour party was ………………, ……………… etc.

Answers:

1. 1939, 1945
2. Versailles
3. 100
4. 1929
5. Nazis
6. Chamberlin
7. Czechoslovakia
8. Jews
9. 1933, 1945
10. modem
11. 1939
12. 1940
13. Blitz
14. Lightning strike
15. Royal Navy
16. France
17. Spitfires, Hurricanes
18. President Roosevelt
19. Oil fields
20. Italian
21. Stalingrad
22. 1943
23. General Elsenhower
24. Guangzhou
25. Japan
26. Solomon
27. Hiroshima, Nagasaki
28. 15th August 1945
29. 2nd Sept 1945
30. United States, Soviet Russia
31. Nuclear weapons
32. United Nations, World Bank, International Monetary Fund
33. Women
34. India
35. Holocaust
36. Israel
37. Israel
38. Human Rights
39. 10th Dec 1948
40. 10th Dec 1948
41. Atlantic Charter
42. Germany, Italy, Japan
43. 51
44. June 26, 1945
45. One
46. Government
47. Executive, Judicial, Legislative, Co-ordinating
48. The right to block major decisions
49. Permanent members
50. Five
51. World Health Organisation
52. United Nations Children’s Fund
53. Food and Agricultural Organisation
54. UN educational, Scientific and Cultural Organisation
55. United Nations Development programme
56. Indian Army
57. Bretton Woods Twins
58. IBRD, IDA
59. International Bank for Reconstruction and Development
60. International Development Agency
61. Government
62. Soft loans
63. 50
64. IFC
65. International Finance Corporation
66. Environment, AIDS
67. Hary Dexter, John Maynard Keynes
68. 29
69. Financial Stability
70. Balance of payment
71. 189
72. Welfare states
73. Labour party
74. National Health Service
75. Old age pension, Child care services

III. Choose the correct statement

Question 1.
(i) The axis powers of World War II were Germany, Italy and Japan.
(ii) Russia attacked the American naval base at Pearl Harbour in 1941.
(iii) The UN adopted the historic human rights charter on 10th December 1947.
(iv) The executive wing of the UN is the UN Secretariat.
(a) (i) (ii) are correct
(b) (i) (ii) (iii) are correct
(c) (i) (iv) are correct
(d) (ii) (iv) are wrong.
Answer:
(c) (i) (iv) are correct

Question 2.
(i) Reparations refers to the compensation exacted from a defeated nation by the victorious nation.
(ii) Slaughter is compulsory military service.
(iii) Japanese navy was defeated by the US Navy at the battle of mid way.
(iv) Progression taxation by taxing the higher income groups at relatively high rates.
(a) (ii) (iii) (iv) are correct
(b) (ii) (iv) are wrong
(c) (iii) (iv) are correct
(d) (iii) (iv) are wrong.
Answer:
(a) (ii) (iii) (iv) are correct

Question 3.
(i) The Security Council of the UNO has fifteen members.
(ii) The mass killing of Jews in Nazi was called holocaust
(iii) Battle of Ex Alamein was considered one of the greatest battles by Russia.
(iv) The Japanese navy defeated the US navy in the battle of Midway.
(a) (i) (ii) are correct
(b) (ii) (iv) are correct
(c) (i) (ii) (iii) are correct
(d) (i) (iv) are wrong
Answer:
(a) (i) (ii) are correct

Question 4.
(i) The World Bank and the IMF are referred to as the Bretton Woods Twins.
(ii) The post World War I led to the rise of dictatorship in Italy, Germany, and Spain.
(iii) The post World War II changed the world into two blocks as communist and non communist.
(iv) The Shakespeare’s play the Merchant of Venice clearlv depicts the dislike and distrust of Jews among the Nazi people.
(a) (i) (ii) are correct
(b) (i) (ii) (iii) are wrong
(c) (i) (ii) (iii) (iv) are correct
(d) (iii) (iv) (ii) are wrong
Answer:
(c) (i) (ii) (iii) (iv) are correct

Question 5.
(i) IMF lends money from its resources to countries facing balance of payment problem.
(ii) The Munich pact was signed between Germany and the Soviet Union.
(iii) Huge worthless money for bread often refers to the Britain’s severe inflation after II World War.
(iv) Franco of Spain was the only dictatorship that emerged after II World War.
(a) (i) (ii) are correct
(b) (i) (iii) (iv) are correct
(c) (i) (ii) (iii) (iv) are wrong
(d) (i) (iv) are correct
Answer:
(a) (i) (ii) are correct

Question 6.
(i) Japanese extended their empire throughout South-east Asia.
(ii) Burma, Indonesia, Singapore, Malaya, Hong Kong, Philippines all fell to the Japanese.
(iii) Many Indians walked ail the way from Burma to the Indian border facing many sufferings.
(iv) Many Indians who stayed there suffered under the Japanese.
(a) (ii) (iv) are wrong
(b) (i) (ii) (iii) (iv) are correct
(c) (iii) (i) are correct
(d) (i) (ii) (iii) (iv) are wrong
Answer:
(b) (i) (ii) (iii) (iv) are correct

Question 7.
(i) Hitler was killed by his countrymen in 1945
(ii) Mussolini committed suicide in April 1945
(iii) The United States declared war on Japan on December 1944.
(iv) In 1938, Japan invaded China and seized Beijing.
(a) (ii) (iv) are wrong
(b) (i) (ii) (iii) (iv) are correct
(c) (i) (ii) are correct
(d) (i) (ii) (iii) (iv) are wrong
Answer:
(d) (i) (ii) (iii) (iv) are wrong

Question 8.
(i) In the year 1940, the British Prime Minister Chamberlain resigned.
(ii) The newly elected British Prime Minister next was Winston Churchill.
(iii) The end of World War II signalled a change in the world order and political configurations among the major powers.
(iv) The Treaty of Versailles ended the World War II.
(a) (i) (iv) are wrong
(b) (i) (ii) (iii) (iv) are correct
(c) (i) (ii) (iii) are correct
(d) (i) (ii) (iv) are correct
Answer:
(c) (i) (ii) (iii) are correct

IV. Assertion and Reason

Question 1.
Assertion (A): World War I (1914-18) and World War II (1939-45) are only referred as World wars.
Reason (R): The high death of the civilians and the soldiers and the extended area of the conflicts.
(a) Both A and R are correct
(b) A is correct but R is not the correct reason
(c) Both A and R are wrong
(d) R is correct but A is wrong.
Answer:
(a) Both A and R are correct

Question 2.
Assertion (A): The League remained an ineffectual international body. Reason (R): Along with the USA, as a non-member mainly Germany was determined to maintain a non-interventionist attitude.
(a) Both A and R are correct
(b) A is correct but R is not the correct reason
(c) Both A and R are wrong
(d) R is correct but A is wrong.
Answer:
(b) A is correct but R is not the correct reason

Question 3.
Assertion (A): Hitler invaded Austria and Czechoslovakia in 1938.
Reason (R): Hitler claimed all the German speaking people should be united into one nation.
(a) Both A and R are correct
(b) A is correct but R is not the correct reason
(c) Both A and R are wrong
(d) R is correct but A is not relevant to R.
Answer:
(a) Both A and R are correct

Question 4.
Assertion (A): The mood in Britain was not in favour of starting another war after World War I.
Reason (R): Just as the United States they wanted to be concerned with the revival of the economy after great depression.
(a) Both A and R are correct
(b) A is correct but R is not the correct reason
(c) Both A and R are wrong
(d) R is correct but A is not relevant to A.
Answer:
(b) A is correct but R is not the correct reason

Question 5.
Assertion (A): Germany developed a fleet of sub-marines which caused havoc in the Atlantic Ocean.
Reason (R): Germany ensured themselves for a sea-borne invasion on allies.
(a) Both A and R are correct
(b) A is right but R is not the correct explanation of A.
(c) Both A and R are wrong
(d) R is correct, which is not relevant to A.
Answer:
(b) A is right but R is not the correct explanation of A.

Question 6.
Assertion (A): The long term objective of Germany was to exploit Russia’s natural Resource oil.
Reason (R): German army invaded Russia.
(a) Both A and R are correct
(b) A is right but R is not the correct explanation of A
(c) Both A and R are wrong
(d) R is correct, which is not relevant to A.
Answer:
(a) Both A and R are correct

Question 7.
Assertion (A): Germans tried to capture the city of Stalingrad in Russia. Reason (R): Stalingrad was the militarised zone of Russia.
(a) Both A and R are correct
(b) A is right, but R is not the correct explanation of A
(c) Both A and R are wrong
(d) R is right, A is wrong.
Answer:
(b) A is right, but R is not the correct explanation of A

Question 8.
Assertion (A): Hitler committed suicide in April 1945.
Reason (R): The Allied forces in 1945, occupied parts of Berlin and began to attack Germany from the east.
(a) Both A and R are correct
(b) A is correct but R is not the correct explanation.
(c) Both A and R are wrong
(d) R is correct but A is not relevant to R.
Answer:
(a) Both A and R are correct

Question 9.
Assertion (A): The United States declared war on Japan.
Reason (R): In 1931, the Japanese army invaded Manchuria and in 1937, invaded China and seized Beijing.
(a) Both A and R are correct
(b) A is correct but R is not the correct explanation of A
(c) Both A and R are wrong
(d) R is correct but R is not relevant to A.
Answer:
(b) A is correct but R is not the correct explanation of A

Question 10.
Assertion (A): U.S. dropped an atomic bomb on Hiroshima and another bomb was dropped on Nagasaki.
Reason (R): U.S. developed hatred over the development of two cities Hiroshima and Nagasaki.
(a) Both A and R are correct
(b) A is correct, but R is not the correct explanation of A
(c) Both A and R are wrong
(d) R is correct but it is the correct reason for A.
Answer:
(b) A is correct, but R is not the correct explanation of A

Question 11.
Assertion (A): The U.S. and the Soviet Union followed communist and non¬communist ideas.
Reason (R): Countries began to devote large amount of resources in developing dangerous weapons.
(a) Both A and R are correct
(b) A is correct, but not relevant to R
(c) Both A and R are wrong
(d) R is correct, but not relevant to A.
Answer:
(c) Both A and R are wrong

V. Match the Following

Question 1.
Match the Column I with Column II.

Answer:
A. (iv)
B. (i)
C. (v)
D. (vi)
E. (iii)

Question 2.
Match the Column I with Column II.

Answer:
A. (v)
B. (i)
C. (vi)
D. (iii)
E. (ii)

VI. Answer the questions briefly

Question 1.
What are soft loans?
Answer:
The loans that are sanctioned by the International Development Agency to the Governments for developmental activities are called as soft loans. They are given at very low rate of interest of 50 years.

Question 2.
Did Munich Pact bring peace for some time? How?
Answer:

1. In September 1938, Hitler threatened Czechoslovakia.
2. The British Prime Minister Neville chamberlain initiated talks and signed Munich pact,
3. Hitler promised not to take any more Czech territory.
4. Chamberlain believed that he had achieved “Peace for some time”. But within six months Hitler seized the remainder of Czechoslovakia. So Munich pact has brought peace only for some time.

Question 3.
What do you know about the World Bank?
Answer:
The World bank consists of two main organs namely The International Bank for Reconstruction and Development (IBRD) and the International Development Agency (IDA). Together they are called as the World Bank.

Question 4.
Why did America declare war on Japan?
Answer:

1. The Japanese had attacked the American fleet stationed at Pearl Harbour on December 7, 1941.
2. This disastrous attack forced the Americans to enter into the war.
3. The very next day the USA declared war on Japan.

Question 5.
What are the axis powers and the ally powers of II World War?
Answer:
Germany, Italy, Japan – Axis powers.
Britain, France, Russia, USA – Ally powers.

Question 6.
Name the countries involved in World War II.
Answer:

1. The allies countries were under the leadership of Britain. [Britain, France, Russia and U.S.A]
2. The axis countries were under the leadership of Germany. [Germany, Italy and Japan]

Question 7.
What was the immediate cause of the II World War?
Answer:
The main and immediate cause of the II World War was the aggressive, military, dictatorship attitude of Germany, fast-developing Japan. Hitler’s attack on Poland in 1939, resulted in the declaration of the War by Britain and France.

Question 8.
Write a brief note on security council.
Answer:

1. The council has five permanent members and ten non-permanent members.
2. The five permanent members are the USA, UK, France, Russian Federation and China,
3. The non-permanent members are elected by the General Assembly for two years term,
4. The permanent members have the right to veto for any council decision.
5. Its main responsibility is to maintain International peace and security.

Question 9.
What is ECOSOC? What are its organs?
Answer:
The Economic and Social Council, is the UN organ which is responsible for co-ordinating all the economic and social work of the United Nations. The Regional Economic commissions functioning for regional development across the various regions of the World are its organs. (Asia pacific, West Asia, Europe, Africa, Latin America).

Question 10.
Name some of the specialized agencies of the UNO.
Answer:

1. The World Health Organisation [WHO]
2. The United Nations Educational Scientific and Cultural Organisation(UNESCO)
3. The United Nation’s Children’s Fund (UNICEF)
4. The International Labour Organisation (ILO)
5. Food and Agricultural Organisation (FAO)
6. International Monetary Fund (IMF)
7. International Bank for Reconstruction and Development (IBRD)

VII. Answer the questions given under each caption

Question 1.
Causes of the Second World War.

(a) Name the treaty signed by Japan, Italy and Germany.
Answer:
Italy – Germany – Japan signed the Rome – Berlin – Tokyo Axis treaty.

(b) Mention some of the ideologies that emerged after the First World War.
Answer:
Democracy, Communism, Fascism and Nazism.

(c) What was the policy followed by the statesmen of the major world powers?
Answer:
The statesmen of the major world powers followed the policy of appeasement.

(d) What did Hitler violate?
Answer:
He violated the Munich Pact.

Question 2.
Munich Pact

(a) Who concluded the Munich pact with Germany?
Answer:
In 1938, Prime Minister Chamberlain concluded the Munich pact with Germany.

(b) What did Hitler do in 1939?
Answer:
In 1939, Hitler invaded Czechoslovakia breaking the munich pact that Germany would not attack any other country.

(c) Which act of Hitler made Britain and France declare war on Germany?
Answer:
His act of attack on Poland made Britain and France declare war on Germany.

(d) What were the weapons used in World War II?
Answer:
Heavy military equipment such as tanks, sub-marines, battleships, aircraft carriers, fighter planes and bomber planes.

Question 3.
Organs of the UNO

(a) Name the major organs of the UNO.
Answer:

1. The General Assembly
2. The Security Council
3. The Economic and Social Council
4. The Trusteeship Council
5. The International Court of Justice
6. The Secretariat

(b) Who was elected as the President of the UN General Assembly in 1953?
Answer:
Mrs. Vijaya Lakshmi Pandit

(c) What is the function of the Trusteeship Council?
Answer:
The Trusteeship Council looks after certain territories placed under the trusteeship of the UNO.

(d) How is the Secretary-General of the UNO appointed?
Answer:
The Secretary-General is appointed by the General Assembly on the advice of the Security Council.

Question 4.
Birth of Israel.

(a) What is meant by Holocaust?
Answer:
Holocaust refers to the mass killing of jews by the Germans during World War II.

(b) What was the major outcome of the Holocaust?
Answer:
The major outcome of the Holocaust was the creation of the State of Israel as a homeland for the Jews.

(c) What did the Israel occupy?
Answer:
The Israel has occupied large parts of Palestinian homelands.

(d) From whom does Israel get the support from?
Answer:
Israel get the vast support from the United States.

Question 5.
The United Nations

(a) Who took the first initiative for the formation of the United Nations?
Answer:
The United states and the Britain in 1941.

(b) Name the joint declaration they issued?
Answer:
The Atlantic Charter was the name of the joint declaration they issued.

(c) How many countries accepted the declaration at first?
Answer:
The declaration of the United Nations was accepted by 26 countries, on New years Day 1942.

(d) How many nations signed the charter? When?
Answer:
On June 26, 1945, 51 nations signed the charter.

Question 6.
International Monetary Fund (IMF)

(a) On whose idea the International Monetary Fund was initiated?
Answer:
Harry Dexter white and John Maynard Keynes ideas brought the emergence of IMF.

(b) When was it formally organised?
Answer:
It was formally organised in 1945 with 29 member countries (at present 189).

(c) What were the three main agendas of the IMF?
Answer:

1. To promote International monetary co-operation.
2. To expand International trade.
3. To bring exchange stability.

(d) State the main reason of its funding?
Answer:
The IMF lends money from its resources to countries facing Balance of payment problems.

VIII. Answer in detail

Question 1.
What were the results of Second World War?
Answer:

1. The destruction to life and property was on a much larger scale than the First World War.
2. Over 50 millions lost their lives.
3. It sounded the death knell to dictatorship in Germany and Italy.
4. Germany was occupied by the Allied forces, and later it was divided into two parts.
5. The West Germany was controlled by Britain, France and America and the East Germany by Russia.
6. At the end of the war, Japan was occupied by American forces under General Mc. Arthur.
7. The war weakened Britain and France.
8. America and Russia emerged as super powers.
9. The war did not end totalitarianism in Russia. A cold war started between Russia and America.
10. The war quickened the phase of national movements in Asia and Africa.
11. India, Burma, Egypt, Ceylon and Malaya won their freedom from Britain.
12. Philippines got independence from America.
13. Indo-china got independence from France.
14. Indonesia got independence from the Dutch.
15. The European countries gave up the policy of colonialism and imperialism.
16. The United Nations Organisation was set up to maintain international peace and harmony. It works hard to maintain international co-operation and for the promotion of human welfare.

Question 2.
Write a note on international Monetary Fund (IMF).
Answer:

1. International Monetary Fund was established in 1945 after the Bretton Woods conference in 1944 along with the World Bank.
2. It is located in the Washington in United States.
3. The idea of starting of IMF was given by Harry Dexter, White and John Maynard Keynes, a famous economist.
4. The initial members of the IMF were 29. Now, there are 189 member countries with IMF,
5. The main objectives of IMF include to foster global monetary co-operation, to secure financial stability, to facilitate trade, promote employment, to sustain economic growth and reduce poverty all over the world.
6. The fund lends money to its member countries to correct their balance of payment position if they are unable to pay for their imports.
7. The funding from IMF is not very easy as it strictly imposes restrictions on lending.
8. It imposes the developing nations to tighten the budgets and reduce fiscal expenditure.

Question 3.
Write a note on the UN Commission of Human Rights.
Answer:

1. Human Rights means the fundamental freedom for all human beings without any differences in race, sex, language and religion.
2. The UN efforts to protect human rights on a global basis resulted in the formation of the UN commission on Human Rights.
3. A committee was set up for its formation. It was headed by the wife of FDR of USA, after his death.
4. The other members of the commission included Charles Malik from Lebanon, P.C.Chang from China, Rene Casin from France.
5. The Commission set forth with 30 articles.
6. The UN adopted this historic charter on 10th December 1948.
7. This day, the 10th December is observed as Human Rights Day all over the World.
8. According to the Franklin and Eleanor institute in New York, reports, from 1948 till now, nearly 90 National constitutions are part of this Human Rights Commission of UN.

Students can download Maths Chapter 6 Trigonometry Ex 6.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.5

Multiple Choice Questions

Question 1.
The value of sin2 θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to ………………
(1) tan² θ
(2) 1
(3) cot² θ
(4) 0
Answer:
(2) 1
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1

Question 2.
tan θ cosec² θ – tan θ is equal to ………………
(1) sec θ
(2) cot² θ
(3) sin θ
(4) cot θ
Answer:
(4) cot θ
Hint:
tan θ cosec² θ – tan θ = tan θ (cosec² θ – 1)
= tan θ × cot² θ = \(\frac{1}{\cot \theta}\) × cot² θ = cot θ

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
(sin α + cos α)² + (cos α + sec α)²
= sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α
= 1 + cosec² α + 2 + sec² α + 2
= 1 + cot² α + 1 + 2 + tan² α + 1 + 2
= 7 + tan² α + cot² α
k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b (a2 – 1) is equal to ……………
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a
Hint:
b (a² – 1) = (sec θ + cosec θ) [(sin θ + cos θ)² – 1]
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\) [sin² θ + cos² θ + 2 sin θ cos θ – 1]

Question 5.
If 5x = sec θ and \(\frac { 5 }{ x } \) = tan θ, then x2 – \(\frac{1}{x^{2}}\) is equal to …………….
(1) 25
(2) \(\frac { 1 }{ 25 } \)
(3) 5
(4) 1
Answer:
(2) \(\frac { 1 }{ 25 } \)
Hint:



Question 6.
If sin θ = cos θ , then 2 tan² θ + sin² θ – 1 is equal to ………………
(1) \(\frac { -3 }{ 2 } \)
(2) \(\frac { 3 }{ 2 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { -2 }{ 3 } \)
Answer:
(2) \(\frac { 3 }{ 2 } \)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then …………..
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:
x = a tan θ
\(\frac { x }{ a } \) = tan θ
\(\frac{x^{2}}{a^{2}}\) = tan² θ
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = sec² θ – tan² θ = 1
y = b sec θ
\(\frac{y}{b}\) = sec θ
\(\frac{y^{2}}{b^{2}}\) = sec² θ

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to ……………
(1) 0
(2) 1
(3) 2
(4) -1
Answer:
(3) 2
Hint:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b-a
Solution:
(2) b² – a²
(a cot θ + b cosec θ)² = p²
(b cot θ + a cosec θ )² = q²
p² – q² = a² cost²θ + a² cot²θ + 2ab cot θ cosec θ – (b²cot²θ + a² cosec²θ + 2ab cot θ cosec θ) = (a² – b²) cot²θ + (b² – a²)cosec²θ = (a² – b²) (cosec²θ – 1) + (b² – a²) (cosec²θ)
= (a² – b²)cosec²θ – (a² – b²) – (a² – b²) cosec²θ
= b² – a²

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt { 3 }\) : 1, then the angle of elevation of the sun has a measure
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°
Hint:
Ratio of length of the tower : length of the shadow = \(\sqrt { 3 }\) : 1

Let the tower be \(\sqrt { 3 }\) x and the shadow be x
tan C = \(\frac { AB }{ BC } \) ⇒ tan C = \(\frac{\sqrt{3} x}{x}\) = \(\sqrt { 3 }\)
tan C = tan 60° ⇒ ∴ ∠C = 60°

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘6’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to ……………
(1) \(\sqrt { 3 }\) b
(2) \(\frac { b }{ 3 } \)
(3) \(\frac { b }{ 2 } \)
(4) \(\frac{b}{\sqrt{3}}\)
Answer:
(3) \(\frac { b }{ 2 } \)
Hint:
Let the height of the pole BC be h
AC = b + h
Let CD be x
In the right ∆ BCD, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ x } \)
x = \(\sqrt { 3 }\) h ………. (1)

In the right ∆ ACD, tan 60° = \(\frac { AC }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { b+h }{ x } \)
x = \(\frac{b+h}{\sqrt{3}}\) ………(2)
From (1) and (2) we get
\(\sqrt { 3 }\) h = \(\frac{b+h}{\sqrt{3}}\) ⇒ 3 h = b + h
2 h = b ⇒ h = \(\frac { b }{ 2 } \)

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to
(1) 41. 92 m
(2) 43. 92 m
(3) 43 m
(4) 45. 6 m
Answer:
(2) 43. 92 m
Hint:
In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \) = \(\frac { 60 }{ x+y } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 60 }{ x+y } \) ⇒ x + y = 60 \(\sqrt { 3 }\)

y = 60 \(\sqrt { 3 }\) – x …….(1)
In the right ∆ ABD, tan 45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 60 }{ y } \) ⇒ y = 60 ………..(2)
From (1) and (2) we get
60 = 60 \(\sqrt { 3 }\) – x
x = 60 \(\sqrt { 3 }\) – 60 = 60 (\(\sqrt { 3 }\) – 1) = 60 (1.732 – 1)
= 60 × 0.732
x = 43.92 m

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is …………….
(1) 20,10\(\sqrt { 3 }\)
(2) 30, 5 \(\sqrt { 3 }\)
(3) 20, 10
(4) 30, 10\(\sqrt { 3 }\)
Answer:
(4) 30, 10\(\sqrt { 3 }\)
Hint:
Let the height of the multistoried building AB be “h”
AE = h – 20
Let BC be x
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \) ⇒ \(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ………..(1)

In the right ∆ ABC, tan 30° = \(\frac { AE }{ ED } \) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
x = (h – 20) \(\sqrt { 3 }\) ………(2)
From (1) and (2) we get,
\(\frac{h}{\sqrt{3}}\) = (h – 20) \(\sqrt { 3 }\)
h = 3h – 60 ⇒ 60 = 2 h
h = \(\frac { 60 }{ 2 } \) = 30
Distance between the building (x) = \(\frac{h}{\sqrt{3}}=\frac{30}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}=10 \sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is ……………….
(1) \(\sqrt { 2 }\)x
(2) \(\frac{x}{2 \sqrt{2}}\)
(3) \(\frac{x}{\sqrt{2}}\)
(4) 2 x
Answer:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:
Consider the height of the 2^nd person ED be “h”
Height of the second person is 2 h
C is the mid point of BD
In the right ∆ ABC, tan θ = \(\frac { AB }{ BC } \)

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β . The angle of depression of its reflection in the lake is 45° . The height of the location of the cloud from the lake is ………….

Students can download Maths Chapter 6 Trigonometry Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10 \(\sqrt { 3 }\) m.
Answer:
Height of the tower (AC) = 10 \(\sqrt { 3 }\) m
Distance between the base of the tower and point of observation (AB) = 30 m
Let the angle of elevation ∠ABC be θ
In the right ∆ ABC, tan θ = \(\frac { AC }{ AB } \)
= \(\frac{10 \sqrt{3}}{30}=\frac{\sqrt{3}}{3}\)

tan θ = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle of inclination is 30°

Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt { 3 }\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30° . Find the width of the road.
Answer:
Let the mid point of the road AB is “P” (PA = PB)
Height of the home = 4\(\sqrt { 3 }\) m
Let the distance between the pedestrian and the house be “x”
In the right ∆ APD, tan 30° = \(\frac { AD }{ AP } \)

\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x}\)
x = 4 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 12 m
∴ Width of the road = PA + PB
= 12 + 12
= 24 m

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the window FE be “h” m
Let FC be “x” m
∴ EC = (h + x) m

In the right ∆ CDF, tan 45° = \(\frac { CE }{ CD } \)
1 = \(\frac { x }{ 5 } \) ⇒ x = 5
In the right ∆ CDE, tan 60° = \(\frac { CE }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { x+h }{ 5 } \) ⇒ x + h = 5\(\sqrt { 3 }\)
5 + h = 5 \(\sqrt { 3 }\) (substitute the value of x)
h = 5 \(\sqrt { 3 }\) – 5 = 5 × 1.732 – 5 = 8. 66 – 5 = 3.66
∴ Height of the window = 3.66 m

Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40° . Find the height of the pedestal.
(tan 40° = 0.8391, \(\sqrt { 3 }\) = 1.732)
Answer:
Height of the statue = 1.6 m
Let the height of the pedestal be “h”
AD = H + 1.6m
Let AB be x
In the right ∆ ABD, tan 60° = \(\frac { AD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h+1.6 }{ x } \)
x = \(\frac{h+1.6}{\sqrt{3}}\) ……..(1)

In the right ∆ ABC, tan 40° = \(\frac { AC }{ AB } \)
0.8391 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.8391 } \)
Substitute the value of x in (1)
\(\frac{h}{0.8391}=\frac{h+1.6}{\sqrt{3}}\)
(h + 1.6) 0.8391 = \(\sqrt { 3 }\) h
0.8391 h + 1.34 = 1.732 h
1.34 = 1.732 h – 0.8391 h
1.34 = 0.89 h
h = \(\frac { 1.34 }{ 0.89 } \) = \(\frac { 134 }{ 89 } \) = 1.5 m
Height of the pedestal = 1.5 m

Question 5.
A Flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30° . Find (i) the height of the pole (ii) radius of the dome. (\(\sqrt { 3 }\) = 1.732)

Answer:
Height of the Flag pole (ED) = h m
AF and AD is the radius of the semi circle (r)
AC = (r + 7)
AB = (r + 7 + 5)
= (r + 12)
In the right ∆ ABD, tan 30° = \(\frac { AD }{ AB } \)

\(\frac{1}{\sqrt{3}}=\frac{r}{r+12}\)
\(\sqrt { 3 }\) r = r + 12
\(\sqrt { 3 }\) r – r = 12 ⇒ r (\(\sqrt { 3 }\) – 1) = 12
r[1.732 – 1] = 12 ⇒ 0.732 r = 12
r = \(\frac { 12 }{ 0.732 } \) ⇒ = 16.39 m
In the right ∆ ACE, tan 45° = \(\frac { AE }{ AC } \)
1 + \(\frac { r+h }{ r+7 } \)
r + 7 = r + h
∴ h = 7 m
Height of the pole (h) = 7 m
Radius of the dome (r) = 16.39 m

Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer:
Let the height of the electric pole AD be “h” m
EC = 15 – h m
Let AB be “x”

In the right ∆ ABC, tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 15 }{ x } \)
x = \(=\frac{15}{\sqrt{3}}=\frac{15 \times \sqrt{3}}{3}\)
= 5\(\sqrt { 3 }\)
In the right ∆ CDE, tan 30° = \(\frac { EC }{ DE } \)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{x}\) ………….(1)
Substitute the value of x = 5 \(\sqrt { 3 }\) in (1)
\(\frac{1}{\sqrt{3}}=\frac{15-h}{5 \sqrt{3}} \Rightarrow \sqrt{3}(15-h)=5 \sqrt{3}\)
(15 – h) = \(\frac{5 \sqrt{3}}{\sqrt{3}}\) ⇒ 15 – h = 5
h = 15 – 5 = 10
∴ Height of the electric pole = 10 m

Question 7.
A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Answer:
Let the first part of the pole be “x” and the second part be “9x”
∴ height of the pole (AC) = x + 9x = 10x
Given ∠CDB = ∠BDA
∴ BD is the angle bisector of ∠ADC
By angle bisector theorem

\(\frac { AB }{ BC } \) = \(\frac { AD }{ DC } \)
\(\frac { 9x }{ x } \) = \(\frac { AD }{ 25 } \) ⇒ AD = 9 × 25 = 225
In the right ∆ ACD
AD² = AC² + CD²
(225)² = (10x)² + 25²
50625 = 100x² + 625
∴ 100x² = 50625 – 625 = 50000
x² = \(\frac { 50000 }{ 100 } \) = 500
x = \(\sqrt { 500 }\) = \(\sqrt{5 \times 100}=10 \sqrt{5}\)
∴ AC = 10 × 10\(\sqrt { 5 }\) = 100 \(\sqrt { 5 }\) (AC = 10x)
∴ Height of the pole = 100 \(\sqrt { 5 }\) m

Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405)
Answer:
Let the height of the peak be “h” mile. Let AD be x mile.
∴ AB = (x + 1) mile.
In the right ∆ ADC, tan 8° = \(\frac { AC }{ AC } \)

0.1405 = \(\frac { h }{ x } \)
x = \(\frac { h }{ 0.1405 } \) ………..(1)
In ∆ ABC, tan 4° = \(\frac { AC }{ AB } \)
0.0699 = \(\frac { h }{ x+1 } \) ⇒ (x + 1) 0.0699 = h
0.0699x + 0.0699 = h
0.0699 x = h – 0.0699
x = \(\frac { h-0.0699 }{ 0.0699 } \) ………(2)
Equation (1) and (2) we get,
\(\frac { h-0.0699 }{ 0.0699 } \) = \(\frac { h }{ 0.1405 } \)
0.0699 h = 0.1405 (h – 0.0699)
0.0699 h = 0.1405 h – 0.0098
0.0098 = 0.1405 h – 0.0699 h
0.0098 = 0.0706 h
h = \(\frac { 0.0098 }{ 0.0706 } \) = \(\frac { 98 }{ 706 } \) = 0.1388
= 0.14 mile (approximately)
Height of the peak = 0.14 mile

Students can download Maths Chapter 6 Trigonometry Ex 6.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.4

Question 1.
From the top of a tree of height 13 m the angle of elevation and depression of the top and bottom of another tree are 45° and 30° respectively. Find the height of the second tree, (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the second tree be “h”
ED = (h – 13) m
Let AB = x m
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 13 }{ x } \)
x = 13 \(\sqrt { 3 }\) ……..(1)

In the right ∆ CED, tan 45° = \(\frac { DE }{ EC } \)
1 = \(\frac { h-13 }{ x } \)
x = h – 13 ……..(2)
From (1) and (2) we get
h – 13 = 13 \(\sqrt { 3 }\) ⇒ h = 13 \(\sqrt { 3 }\) + 13
= 13 × 1.732 + 13
= 22.52 + 13 = 35.52 m
∴ Height of the second tree = 35.52 m

Question 2.
A man is standing on the deck of a ship, which is 40 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30° . Calculate the distance of the hill from the ship and the height of the hill. (\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the hill BE be “h” m and the distance of the hill from the ship be “x” m
In the right ∆ ABD
tan 30° = \(\frac { AD }{ DB } \)
\(\frac{1}{\sqrt{3}}=\frac{40}{x}\)
x = 40 \(\sqrt { 3 }\) ……..(1)
In the right ∆ CDE
tan 60° = \(\frac { CE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-40 }{ x } \)
x = \(\frac{h-40}{\sqrt{3}}\) ……..(2)

From (1) and (2) we get
\(\frac{h-40}{\sqrt{3}}\) = 40\(\sqrt { 3 }\)
h – 40 = 40 × 3
h = 120 + 40 = 160 m
Height of the hill = 160 m
Distance of the hill from the ship = 40 × \(\sqrt { 3 }\) = 40 × 1.732 = 69.28 m

Question 3.
If the angle of elevation of a cloud from a point ‘h’ metres above a lake is θ1 and the angle of depression of its reflection in the lake is θ2. Prove that the height that the cloud is located from the ground is \(\frac{h\left(\tan \theta_{1}+\tan \theta_{2}\right)}{\tan \theta_{2}-\tan \theta_{1}}\)
Answer:
Let P be the cloud and Q be its reflection.
Let A be the point of observation such that AB = h
Let the height of the cloud be x. (PS = x)
PR = x – h and QR = x + h

Let AR = y
In the right ∆ ARP, tan θ₁ = \(\frac { PR }{ AR } \)
tan θ₁ = \(\frac { x-h }{ y } \) ………(1)
In the ∆ AQR,
tan θ₂ = \(\frac { QR }{ AR } \)
tan θ₂ = \(\frac { x+h }{ y } \) ……….(2)
Add (1) and (2)

Question 4.
The angle of elevation of the top of a cell phone tower from the foot of a high apartment is 60° and the angle of depression of the foot of the tower from the top of the apartment is 30° . If the height of the apartment is 50 m, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be 120 m. State if the height of the above mentioned cell phone tower meets the radiation norms.
Answer:
Let the height of the cell phone tower be “h” m
AD is the height of the apartment; AD = 50 m
Let AB be “x”
In the right triangle ABC
tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)

x = \(\frac{h}{\sqrt{3}}\) …….(1)
In the right triangle ABD, tan 30° = \(\frac { AD }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 50 }{ x } \)
x = 50 \(\sqrt { 3 }\) ……(2)
From (1) and (2) We get
\(\frac{h}{\sqrt{3}}\) = 50 \(\sqrt { 3 }\)
h = 50\(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3 = 150
Height of the cell phone tower is 150 m.
Yes, the cell phone tower meets the radiation norms.

Question 5.
The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. (\(\sqrt { 3 }\) = 1.732)
Answer:
(i) Let the height of the lamp post AE be “h” m
DE = h – 66
Let AB be “x”
In the right ∆ ABC, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}=\frac{66}{x}\)
x = 66 \(\sqrt { 3 }\) ……(1)

In the right ∆ CDE, tan 60° = \(\frac { DE }{ DC } \)
\(\sqrt { 3 }\) = \(\frac { h-66 }{ x } \) ⇒ \(\sqrt { 3 }\) x = h – 66
x = \(\frac{h-66}{\sqrt{3}}\) ………….(2)
From (1) and (2) we get
\(\frac{h-66}{\sqrt{3}}\) = 66 \(\sqrt { 3 }\)
h – 66 = 66 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 66 × 3
h – 66 = 198 ⇒ h = 198 + 66
h = 264 m
(i) the height of the lamp post = 264 m
(ii) Difference of the height of lamp post and apartment = 264 – 66
= 198 m
(ii) Distance between the lamp post and the apartment = 66 \(\sqrt { 3 }\) m
= 66 × 1.732 = 114.31 m

Question 6.
Three villagers A, B and C can see each other across a valley. The horizontal distance between A and B is 8 km and the horizontal distance between B and C is 12 km. The angle of depression of B from A is 20° and the angle of elevation of C from B is 30°. Calculate:
(i) the vertical height between A and B.
(ii) the vertical height between B and C. (tan 20° = 0 .3640, \(\sqrt { 3 }\) = 1. 732)
Answer:
Let AD is the vertical height between A and B
In the right ∆ ABD

tan 20° = \(\frac { AD }{ BD } \)
0.3640 = \(\frac { AD }{ 8 } \)
AD = 0.3640 × 8 = 2.912 km
∴ AD = 2.91 km
CE is the vertical height between C and B
In the right ∆ BCE, tan 30° = \(\frac { CE }{ BE } \)
\(\frac{1}{\sqrt{3}}=\frac{C E}{12} \Rightarrow \sqrt{3} C E=12\)
CE = \(\frac{12}{\sqrt{3}}=\frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{12 \times \sqrt{3}}{3}\)
= 4 \(\sqrt { 3 }\) = 4 × 1.732 = 6.928
= 6.93 km
(i) The vertical height between A and B = 2.91 km
(ii) The vertical height between B and C = 6.93 km

Students can download Maths Chapter 6 Trigonometry Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 1.
Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
Answer:
(i) L. H. S = cot θ + tan θ
= \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\)
= \(\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}\)
[cos² θ + sin² θ = 1]
= \(\frac{1}{\sin \theta \cos \theta}\)
= sec θ . cosec θ = R. H. S
∴ cot θ + tan θ = sec θ cosec θ

(ii) tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
L.H.S = tan⁴ θ + tan² θ
= tan² θ (tan² θ + 1)
= tan² θ sec² θ
R.H.S = sec⁴ θ – sec² θ
= sec² θ (sec² θ – 1)
= sec² θ tan² θ
L.H.S = R.H.S
∴ tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

Question 2.
Prove the following identities.
(i) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan² θ
(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ
Answer:
(i) \(\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}\) = tan² θ

(ii) \(\frac{\cos \theta}{1+\sin \theta}\) = sec θ – tan θ

Aliter:
L.H.S = \(\frac{\cos \theta}{1-\sin \theta}\)
[conjugate (1 – sin θ)]

Question 3.
Prove the following identities.

Solution:

Question 4.
Prove the following identities.
(i) sec⁶ θ = tan⁶ θ + 3 tan² θ sec² θ + 1
(ii) (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²
Answer:
(i) sec⁶ θ = tan⁶ θ + 3 tan² θ sec² θ + 1
L.H.S = sec⁶ θ
= (sec² θ)³ = (1 + tan² θ)³
= 1 + (tan² θ)³ + 3 (1) (tan² θ) (1 + tan² θ) [(a + b)³ = a³ + b³ + 3 ab (a + b)]
= 1 + tan⁶ θ + 3 tan² θ(1 + tan² θ)
= 1 + tan⁶ θ + 3 tan² θ (sec² θ)
= 1 + tan⁶ θ + 3 tan² θ sec² θ
= tan⁶ θ + 3 tan² θ sec² θ + 1
L.H.S = R.H.S

(ii) (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²
L.H.S = (sin θ + sec θ)² + (cos θ + cosec θ)²]
= [sin² θ + sec² θ + 2 sin θ sec θ + cos² θ + cosec² θ + 2 cos θ cosec θ]
= (sin² θ + cos² θ) + (sec² θ + cosec² θ) + 2 (sin θ sec θ + cos θ cosec θ)

= 1 + sec² θ + cosec² θ + 2 sec θ cosec θ
= 1 + (sec θ + cosec θ)²
L.H.S = R.H.S
∴ (sin θ + sec θ)² + (cos θ + cosec θ)² = 1 + (sec θ + cosec θ)²

Question 5.
Prove the following identities.
(i) sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ = 1
(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)
Answer:
(i) L.H.S = sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ

L.H.S = R.H.S
∴ sec⁴ θ (1 – sin⁴ θ) – 2 tan² θ = 1

(ii) \(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\csc \theta-1}{\csc \theta+1}\)

Question 6.
Prove the following identities.


Answer:

Question 7.
(i) If sin θ + cos θ = \(\sqrt { 3 }\), then prove that tan θ + cot θ = 1.
(ii) If \(\sqrt { 3 }\) sin θ – cos θ = θ, then show that tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
Answer:
sin θ + cos θ = \(\sqrt { 3 }\) (squaring on both sides)
(sin θ + cos θ)² = (\(\sqrt { 3 }\))²
sin² θ + cos² θ + 2 sin θ cos θ = 3
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 3 – 1
2 sin θ cos θ = 2
∴ sin θ cos θ = 1
L.H.S = tan θ + cot θ

L.H.S = R.H.S ⇒ tan θ + cot θ = 1

(ii) If \(\sqrt { 3 }\) sin θ – cos θ = 0
To prove tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
\(\sqrt { 3 }\) sin θ – cos θ = 0
\(\sqrt { 3 }\) sin θ = cos θ
\(\frac{\sin \theta}{\cos \theta}=\frac{1}{\sqrt{3}}\)
tan θ = tan 30°
θ = 30°
L.H.S = tan 3θ°
= tan3 (30°)
= tan 90°
= undefined (∝)

∴ tan 3θ = \(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)

Question 8.
(i) If \(\frac{\cos \alpha}{\cos \beta}=m\) and \(\frac{\cos \alpha}{\cos \beta}=n\) then prove that (m2 + n2) cos2 β = n2
(ii) If cot θ + tan θ = x and sec θ – sec θ – cos θ = y, then prove that (x2y)2/3 – (xy2)2/3 = 1
Answer:
(i) L.H.S = (m² + n²) cos² β

L.H.S = R.H.S ⇒ ∴ (m² + n²) cos² β = n²

(ii) Given cot θ + tan θ = x sec θ – cos θ = y
x = cot θ + tan θ
x = \(\frac{1}{\tan \theta}\) + tan θ
= \(\frac{1+\tan ^{2} \theta}{\tan \theta}\) = \(\frac{\sec ^{2} \theta}{\tan \theta}\)

y = sec θ – cos θ
= \(\frac{1}{\cos \theta}-\cos \theta=\frac{1-\cos ^{2} \theta}{\cos \theta}\)
y = \(\frac{\sin ^{2} \theta}{\cos \theta}\)

Question 9.
(i) If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q (p² – 1) = 2 p
(ii) If sin θ (1 + sin² θ) = cos² θ, then prove that cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4
Answer:
(i) p = sin θ + cos θ
p² = (sin θ + cos θ)²
= sin² θ + cos² θ + 2 sin θ cos θ = 1 + 2 sin θ cos θ
q = sec θ + cosec θ
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\)
L.H.S = q(p² – 1)

(ii) sin θ (1 + sin² θ) = cos² θ
sin θ (1 + 1 – cos² θ) = cos² θ
sin θ (2 – cos² θ) = cos² θ
Squaring on both sides,
sin² θ (2 – cos² θ)² = cos⁴ θ
(1 – cos² θ) (4 + cos⁴ θ – 4 cos² θ) = cos⁴ θ
4 cos⁴ θ – 4 cos² θ – cos⁶ θ + 4 cos⁴ θ = cos⁴ θ
4 + 5 cos⁴ θ – 8 cos² θ – cos⁶ θ = cos⁴ θ
– cos⁶ θ + 5 cos⁴ θ – cos⁴ θ – 8 cos² θ = -4
– cos⁶ θ + 4 cos⁴ θ – 8 cos² θ = -4
cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4
Hence it is proved

Question 10.
If \(\frac{\cos \theta}{1+\sin \theta}\) = \(\frac { 1 }{ a } \), then prove that \(\frac{a^{2}-1}{a^{2}+1}\) = sin θ
Answer:

Students can download Maths Chapter 6 Trigonometry Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
From the top of a rock 50 \(\sqrt { 3 }\) m high, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.
Answer:
Let the distance of the car from the rock is “x” m

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \)
\(\frac{1}{\sqrt{3}}=\frac{50 \sqrt{3}}{x}\)
x = 50 \(\sqrt { 3 }\) × \(\sqrt { 3 }\) = 50 × 3
= 150 m
∴ Distance of the car from the rock = 150 m

Question 2.
The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.
Answer:
Let the height of the first building AD be “x” m
∴ EC = 120 – x
In the right ∆ CDE,
tan 45° = \(\frac { CE }{ CD } \)

1 = \(\frac { 120-x }{ 70 } \) ⇒ 70 = 120 – x
x = 50 cm
∴ The height of the first building is 50 m

Question 3.
From the top of the tower 60 m high the anles of depression the top and bottom of a vertical lamp post are observed be 38° and 60° respectively
Find the height of the lamp post. (tan 38° = 0.7813,\(\sqrt { 3 }\) = 1.732)
Answer:
Let the height of the lamp post be “h”
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ ABC,

tan 60° = \(\frac { BC }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { 60 }{ x } \)
x = \(\frac{60}{\sqrt{3}}\) ……..(1)
In the right ∆ DEC, tan 38° = \(\frac { EC }{ DE } \)
0.7813 = \(\frac { 60-h }{ x } \)
x = \(\frac { 60-h }{ 0.7813 } \) …….(2)
From (1) and (2) we get
\(\frac{60}{\sqrt{3}}\) = \(\frac { 60-h }{ 0.7813 } \)
60 × 0.7813 = 60 \(\sqrt { 3 }\) – \(\sqrt { 3 }\) h
\(\sqrt { 3 }\) h = 60 \(\sqrt { 3 }\) – 46.88
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732 h = 57.04 ⇒ h = \(\frac { 57.04 }{ 1.732 } \)
h = \(\frac { 570440 }{ 1732 } \) = 32.93 m
∴ Height of the lamp post = 32.93 m

Question 4.
An aeroplane at an altitude of 1800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are
60° and 30° respectively. Find the distance between the two boats. (\(\sqrt { 3 }\) = 1.732)
Answer:
C and D are the position of the two boats.
Let the distance between the two boats be “x”
Let BC = y
∴ BD = (x + y)

In the right ∆ ABC, tan 30° = \(\frac { AB }{ BD } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 1800 }{ x+y } \)
x + y = 1800 \(\sqrt { 3 }\)
y = 1800 \(\sqrt { 3 }\) – x ……(1)
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \)
\(\sqrt { 3 }\) = \(\frac { 1800 }{ y } \)
y = \(\frac{1800}{\sqrt{3}}\) ……….(2)
From (1) and (2) we get
\(\frac{1800}{\sqrt{3}}\) = 1800 \(\sqrt { 3 }\) – x
1800 = 1800 × 3 – \(\sqrt { 3 }\)x
\(\sqrt { 3 }\)x = 5400 – 1800
x = \(\frac{3600}{\sqrt{3}}=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \times \sqrt{3}}{3}\)
= 1200 × 1.732 = 2078.4 m
Distance between the two boats = 2078.4 m

Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is \(\frac{4 h}{\sqrt{3}}\) m.
Answer:
A and C be the position of two ships.
Let AB be x and BC be y. Distance between the two ships is x + y

In the right ∆ ABD, tan 60° = \(\frac { BD }{ AB } \)
\(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ……(1)
In the right ∆ BCD,
tan 30° = \(\frac { BD }{ BC } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ y } \)
y = \(\sqrt { 3 }\) h
Distance between the two ships (x + y) = \(\frac{h}{\sqrt{3}}+\sqrt{3} h\)
= \(\frac{h+3 h}{\sqrt{3}}=\frac{4 h}{\sqrt{3}}\)
Hence it is verified

Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30 \(\sqrt { 3 }\) feet from the entrance of the lift, find the speed of the lift which is descending.
Answer:
Let the speed of the lift is “x” feet / minute
Distance AB = 2 x feet (speed × time)
BC = (90 – 2x)
In the right ∆ BCD,

tan 30° = \(\frac { BC }{ DC } \)
\(\frac{1}{\sqrt{3}}=\frac{90-2 x}{30 \sqrt{3}}\)
\(\sqrt { 3 }\) (90 – 2x) = 30\(\sqrt { 3 }\)
(90 – 2x) = \(\frac{30 \sqrt{3}}{\sqrt{3}}\) ⇒ (90 – 2x) = 30
2x = 60
x = \(\frac { 60 }{ 2 } \) = 30
x = 30 feet/minute
Speed of the lift = 30 feet / minute (or) [ \(\frac { 30 }{ 60 } \) second) 0.5 feet / second

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Multiple Choice Questions

Question 1.
The area of triangle formed by the points (-5, 0), (0, – 5) and (5, 0) is …………..
(1) 0 sq.units
(2) 25 sq.units
(3) 5 sq.units
(4) none of these
Answer:
(2) 25 sq.units Hint.
Hint:
Area of the ∆

Question 2.
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is …………
(1) x = 10
(2) y = 10
(3) x = 0
(4) y = 0
Answer:
(1) x = 10
Hint:

Question 3.
The straight line given by the equation x = 11 is …………….
(1) parallel to X axis
(2) parallel to Y axis
(3) passing through the origin
(4) passing through the point (0,11)
Answer:
(2) parallel to Y axis

Question 4.
If (5,7), (3,p) and (6,6) are collinear, then the value of p is ……………
(1) 3
(2) 6
(3) 9
(4) 12
Answer:
(3) 9
Hint:
Since the three points are collinear. Area of a triangle is 0

5p + 18 + 42 – (21 + 6p + 30) = 0
5p + 60 – (51 + 6p) = 0
5p + 60 – 51 – 6p = 0
-p + 9 = 0
-p = -9
p = 9

Question 5.
The point of intersection of 3x – y = 4 and x + 7 = 8 is ……………
(1) (5,3)
(2) (2,4)
(3) (3,5)
(4) (4, 4)
Answer:
(3) (3, 5)

Substitute the value of x = 3 in (2)
3 + 7 = 8
y = 8 – 3 = 5
The point of intersection is (3, 5)

Question 6.
The slope of the line joining (12, 3), (4, a) is \(\frac { 1 }{ 8 } \). The value of ‘a’ is …………….
(1) 1
(2) 4
(3) -5
(4) 2
Answer:
(4) 2
Hint:
Slope of a line = \(\frac { 1 }{ 8 } \)

Question 7.
The slope of the line which is perpendicular to a line joining the points (0, 0) and (- 8, 8) is ………..
(1) -1
(2) 1
(3) \(\frac { 1 }{ 3 } \)
(4) -8
Answer:
(2) 1
Hint:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-0 }{ -8-0 } \) = \(\frac { 8 }{ -8 } \) = -1
Slope of the Perpendicular = 1

Question 8.
If slope of the line PQ is \(\frac{1}{\sqrt{3}}\) then slope of the perpendicular bisector of PQ is …………..
(1) \(\sqrt { 3 }\)
(2) –\(\sqrt { 3 }\)
(3) \(\frac{1}{\sqrt{3}}\)
(4) 0
Answer:
(2) –\(\sqrt { 3 }\)
Hint:
Slope of a line = \(\frac{1}{\sqrt{3}}\)
Slope of the ⊥r bisector = –\(\sqrt { 3 }\)

Question 9.
If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is ……………
(1) 8x + 5y = 40
(2) 8x – 5y = 40
(3) x = 8
(4) y = 5
Answer:
(1) 8x + 5y = 40
Hint:
Let the point A be (0, 8) and B (5, 0)

Question 10.
The equation of a line passing through the origin and perpendicular to the line lx -3y + 4 = 0 is
(1) 7x – 3y + 4 = 0
(2) 3x – 7y + 4 = 0
(3) 3x + 7y = 0
(4) 7x – 3y = 0
Answer:
(3) 3x + 7y = 0
Hint:
Slope of the line of 7x – 3y + 4 = 0
= \(\frac { -7 }{ -3 } \) = \(\frac { 7 }{ 3 } \)
Slope of its ⊥^r = \(\frac { -3 }{ 7 } \)
The line passes through (0,0)
Equation of a line is
y – y₁ = m(x – x₁)
y – 0 = \(\frac { -3 }{ 7 } \) (x – 0)
y = \(\frac { -3 }{ 7 } \) x ⇒ 7y = -3x
3x + 7y = 0

Question 11.
Consider four straight lines
(i) l₁ : 3y = 4x + 5
(ii) l₂ : 4y = 3x – 1
(iii) l₃ : 4y + 3x = 7
(iv) l₄ : 4x + 3y = 2
Which of the following statement is true?
(1) l₁ and l₂ are perpendicular
(2) l₂ and l₄ are parallel
(3) l₂ and l₄ are perpendicular
(4) l₂ and l₃ are parallel
Answer:
(3) l₂ and l₄ are perpendicular
Hint:
Slope of l₁ = \(\frac { 4 }{ 3 } \); Slope of l₂ = \(\frac { 3 }{ 4 } \)
Slope of l₃ = – \(\frac { 3 }{ 4 } \); Slope of l₄ = –\(\frac { 4 }{ 3 } \)
(1) l₁ × l₂ = \(\frac { 4 }{ 3 } \) × \(\frac { 3 }{ 4 } \) = 1 …….False
(2) l₁ = \(\frac { 4 }{ 3 } \); l₄ = – \(\frac { 4 }{ 3 } \) not parallel ………False
(3) l₂ × l₄ = \(\frac { 3 }{ 4 } \) × – \(\frac { 4 }{ 3 } \) = -1 …….True
(4) l₂ = \(\frac { 3 }{ 4 } \); l₃ = – \(\frac { 3 }{ 4 } \) not parallel ………False

Question 12.
A straight line has equation 87 = 4x + 21. Which of the following is true …………………….
(1) The slope is 0.5 and the y intercept is 2.6
(2) The slope is 5 and the y intercept is 1.6
(3) The slope is 0.5 and they intercept is 1.6
(4) The slope is 5 and the y intercept is 2.6
Answer:
(1) The slope is 0.5 and they intercept is 2.6
Hint:
8y = 4x + 21
y = \(\frac { 4 }{ 8 } \) x + \(\frac { 21 }{ 8 } \)
= \(\frac { 1 }{ 2 } \) x + \(\frac { 21 }{ 8 } \)
\(\frac { 1 }{ 2 } \) = 0.5
\(\frac { 21 }{ 8 } \) = 2.625
Slope = \(\frac { 1 }{ 2 } \) = 0.5
y intercept = \(\frac { 21 }{ 8 } \) = 2.6

Question 13.
When proving that a quadrilateral is a trapezium, it is necessary to show
(1) Two sides are parallel.
(2) Two parallel and two non-parallel sides.
(3) Opposite sides are parallel.
(4) All sides are of equal length.
Solution:
(2) Two parallel and two non-parallel sides.

Question 14.
When proving that a quadrilateral is a parallelogram by using slopes you must find …………………
(1) The slopes of two sides
(2) The slopes of two pair of opposite sides
(3) The lengths of all sides
(4) Both the lengths and slopes of two sides
Answer:
(2) The slopes of two pair of opposite sides

Question 15.
(2,1) is the point of intersection of two lines.
(1) x – y – 3 = 0; 3x – y – 7 = 0
(2) x + y = 3; 3x + y = 7
(3) 3x + y = 3; x + y = 7
(4) x + 3y – 3 = 0; x – y – 7 = 0
Solution:
(2) x + y = 3; 3x + y = 7

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.4

Question 1.
Find the slope of the following straight lines.
(i) 5y – 3 = 0
(ii) 7x – \(\frac { 3 }{ 17 } \) = 0
Solution:
(i) 5y – 3 = 0
5y = 3 ⇒ y = \(\frac { 3 }{ 5 } \)
Slope = 0

(ii) 7x – \(\frac { 3 }{ 17 } \) = 0 (Comparing with y = mx + c)
7x = \(\frac { 3 }{ 17 } \)
Slope is undefined

Question 2.
Find the slope of the line which is
(i) parallel to y = 0.7x – 11
(ii) perpendicular to the line x = -11
Solution:
(i) y = 0.7x – 11
Slope = 0.7 (Comparing with y = mx + c)
(ii) Perpendicular to the line x = – 11
Slope is undefined (Since the line is intersecting the X-axis)

Question 3.
Check whether the given lines are parallel or perpendicular
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 and \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Solution:
(i) \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) + \(\frac { 1 }{ 7 } \) = 0 ; \(\frac { 2x }{ 3 } \) + \(\frac { y }{ 2 } \) + \(\frac { 1 }{ 10 } \) = 0
Slope of the line (m₁) = \(\frac { -a }{ b } \)
= – \(\frac { 1 }{ 3 } \) ÷ \(\frac { 1 }{ 4 } \) = –\(\frac { 1 }{ 3 } \) × \(\frac { 4 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
Slope of the line (m₂) = – \(\frac { 2 }{ 3 } \) ÷ \(\frac { 1 }{ 2 } \) = –\(\frac { 2 }{ 3 } \) × \(\frac { 2 }{ 1 } \) = – \(\frac { 4 }{ 3 } \)
m₁ = m₂ = – \(\frac { 4 }{ 3 } \)
∴ The two lines are parallel.

(ii) 5x + 23y + 14 = 0 and 23x – 5x + 9 = 0
Slope of the line (m₁) = \(\frac { -5 }{ 23 } \)
Slope of the line (m₂) = \(\frac { -23 }{ -5 } \) = \(\frac { 23 }{ 5 } \)
m₁ × m₂ = \(\frac { -5 }{ 23 } \) × \(\frac { 23 }{ 5 } \) = -1
∴ The two lines are perpendicular

Question 4.
If the straight lines 12y = -(p + 3)x + 12, 12x – 7y = 16 are perpendicular then find ‘p’
Solution:
Slope of the first line 12y = -(p + 3)x +12
y = \(-\frac{(p+3) x}{12}+1\) (Comparing with y = mx + c)
Slope of the second line (m₁) = \(\frac { -(p+3) }{ 12 } \)
Slope of the second line 12x – 7y = 16
(m₂) = \(\frac { -a }{ b } \) = \(\frac { -12 }{ -7 } \) = \(\frac { 12 }{ 7 } \)
Since the two lines are perpendicular
m₁ × m₂ = -1
\(\frac { -(p+3) }{ 12 } \) × \(\frac { 12 }{ 7 } \) = -1 ⇒ \(\frac { -(p+3) }{ 7 } \) = -1
-(p + 3) = -7
– p – 3 = -7 ⇒ -p = -7 + 3
-p = -4 ⇒ p = 4
The value of p = 4

Question 5.
Find the equation of a straight line passing through the point P(-5,2) and parallel to the line joining the points Q(3, -2) and R(-5,4).
Solution:
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line QR = \(\frac { 4+2 }{ -5-3 } \) = \(\frac { 6 }{ -8 } \) = \(\frac { 3 }{ -4 } \) ⇒ – \(\frac { 3 }{ 4 } \)
Slope of its parallel = – \(\frac { 3 }{ 4 } \)
The given point is p(-5, 2)
Equation of the line is y – y₁ = m(x – x₁)
y – 2 = – \(\frac { 3 }{ 4 } \) (x + 5)
4y – 8 = -3x – 15
3x + 4y – 8 + 15 = 0
3x + 4y + 7 = 0
The equation of the line is 3x + 4y + 7 = 0

Question 6.
Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution:
Let the vertices A (6, 7), B (2, -3), D (6, -2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-7 }{ 2-6 } \) = \(\frac { -10 }{ -4 } \) = \(\frac { 5 }{ 2 } \)
Slope of its perpendicular (CD) = – \(\frac { 2 }{ 5 } \)
Equation of the line CD is y – y₁ = m(x – x₁)
y + 2 = –\(\frac { 2 }{ 5 } \) (x – 6)
5(y + 2) = -2 (x – 6)
5y + 10 = -2x + 12
2x + 5y + 10 – 12 = 0
2x + 5y – 2 = 0
The equation of the line is 2x + 5y – 2 = 0

Question 7.
A(-3,0) B(10, -2) and C(12,3) are the vertices of ∆ABC. Find the equation of the altitude through A and B.
Solution:
To find the equation of the altitude from A.
The vertices of ∆ABC are A(-3, 0), B(10, -2) and C(12, 3)

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 3+2 }{ 12-10 } \) = \(\frac { 5 }{ 2 } \)
Slope of the altitude AD is – \(\frac { 2 }{ 5 } \)
Equation of the altitude AD is
y – y₁ = m (x – x₁)
y – 0 = – \(\frac { 2 }{ 5 } \) (x + 3)
5y = -2x -6
2x + 5y + 6 = 0
Equation of the altitude AD is 2x + 5y + 6 = 0
Equation of the altitude from B

Slope of AC = \(\frac { 3-0 }{ 12+3 } \) = \(\frac { 3 }{ 15 } \) = \(\frac { 1 }{ 5 } \)
Slope of the altitude AD is -5
Equation of the altitude BD is y – y₁= m (x – x₁)
7 + 2 = -5 (x – 10)
y + 2 = -5x + 50
5x + 7 + 2 – 50 = 0 ⇒ 5x + 7 – 48 = 0
Equation of the altitude from B is 5x + y – 48 = 0

Question 8.
Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6, -4).
Solution:
“C” is the mid point of AB also CD ⊥ AB.

Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { -4-2 }{ 6+4 } \) = \(\frac { -6 }{ 10 } \) = – \(\frac { 3 }{ 5 } \)
Slope of the ⊥^r AB is \(\frac { 5 }{ 3 } \)
Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { -4+6 }{ 2 } \),\(\frac { 2-4 }{ 2 } \)) = (\(\frac { 2 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (1,-1)
Equation of the perpendicular bisector of CD is
y – y₁ = m(x – x₁)
y + 1 = \(\frac { 5 }{ 3 } \) (x – 1)
5(x – 1) = 3(y + 1)
5x – 5 = 3y + 3
5x – 3y – 5 – 3 = 0
5x – 3y – 8 = 0
Equation of the perpendicular bisector is 5x – 3y – 8 = 0

Question 9.
Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0.
Solution:
Given lines are.

x = \(\frac { 43 }{ 43 } \) = 1
Substitute the value of x = 1 in (1)
7(1) + 3y = 10 ⇒ 3y = 10 – 7
y = \(\frac { 3 }{ 3 } \) = 1
The point of intersection is (1,1)
Equation of the line parallel to 13x + 5y + 12 = 0 is 13x + 5y + k = 0
This line passes through (1,1)
13 (1) + 5 (1) + k = 0
13 + 5 + k = 0 ⇒ 18 + k = 0
k = -18
∴ The equation of the line is 13x + 5y – 18 = 0

Question 10.
Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0.
Solution:
Given lines are.

Substitute the value of x = \(\frac { 16 }{ 7 } \) in (2)
3 × \(\frac { 16 }{ 7 } \) + 2y = 10 ⇒ 2y = 10 – \(\frac { 48 }{ 7 } \)
2y = \(\frac { 70-48 }{ 7 } \) ⇒ 2y = \(\frac { 22 }{ 7 } \)
y = \(\frac{22}{2 \times 7}\) = \(\frac { 11 }{ 7 } \)
The point of intersect is (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
Equation of the line perpendicular to 4x – 7y + 13 = 0 is 7x + 4y + k = 0
This line passes through (\(\frac { 16 }{ 7 } \),\(\frac { 11 }{ 7 } \))
7 (\(\frac { 16 }{ 7 } \)) + 4 (\(\frac { 11 }{ 7 } \)) + k = 0 ⇒ 16 + \(\frac { 44 }{ 7 } \) + k = 0
\(\frac { 112+44 }{ 7 } \) + k = 0 ⇒ \(\frac { 156 }{ 7 } \) + k = 0
k = – \(\frac { 156 }{ 7 } \)
Equation of the line is 7x + 4y – \(\frac { 156 }{ 7 } \) = 0
49x + 28y – 156 = 0

Question 11.
Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y -4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3.
Solution:
The given lines are.


Substitute the value of x = 0 in (1)
3 (0) + y = -2
y = -2
The point of intersection is (0, -2).
The given equation is

Substitute the value of y = \(\frac { 9 }{ 11 } \) in (6)
– x + 2 (\(\frac { 9 }{ 11 } \)) = 3 ⇒ -x + \(\frac { 18 }{ 11 } \) = 3
-x = 3 – \(\frac { 18 }{ 11 } \) = \(\frac { 33-18 }{ 11 } \) = \(\frac { 15 }{ 11 } \)
x = – \(\frac { 15 }{ 11 } \)
The point of intersection is (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \))
Equation of the line joining the points (0, -2) and (-\(\frac { 15 }{ 11 } \),\(\frac { 9 }{ 11 } \)) is


31 × (- 11x) = 11 × 15 (y + 2) = 165 (y + 2)
– 341 x = 165 y + 330
– 341 x – 165 y – 330 = 0
341 x + 165 y + 330 = 0
(÷ by 11) ⇒ 31 x + 15 y + 30 = 0
The required equation is 31 x + 15 y + 30 = 0

Question 12.
Find the equation of a straight line through the point of intersection of the lines 8JC + 3j> = 18, 4JC + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7,6).
Solution:
Given lines are.
8x + 3y = 18 …..(1)
4x + 5y = 9 …..(2)

x = \(\frac { 63 }{ 28 } \) = \(\frac { 9 }{ 4 } \)
Substitute the value of x = \(\frac { 9 }{ 4 } \) in (2)
4 (\(\frac { 9 }{ 4 } \)) + 5y = 9
9 + 5y = 9 ⇒ 5y = 9 – 9
5y = 0 ⇒ y = 0
The point of intersection is (\(\frac { 9 }{ 4 } \),0)
Mid point of the points (5, -4) and (-7, 6)

Equation of the line joining the points (\(\frac { 9 }{ 4 } \),0) and (-1,1)

-13y = 4x – 9
-4x – 13y + 9 = 0 ⇒ 4x + 13y – 9 = 0
The equation of the line is 4x + 13y – 9 = 0

Students can download Maths Chapter 5 Coordinate Geometry Unit Exercise 5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Unit Exercise 5

Question 1.
PQRS is a rectangle formed by joining the points P(- 1, – 1), Q(- 1, 4) , R(5, 4) and S (5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Answer:

Mid point of a line = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
Mid point of PQ (A) = (\(\frac { -1-1 }{ 2 } \),\(\frac { -1+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (-1,\(\frac { 3 }{ 2 } \))
Mid point of QR (B) = (\(\frac { -1+5 }{ 2 } \),\(\frac { 4+4 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { 8 }{ 2 } \)) = (2,4)
Mid point of RS (C) = (\(\frac { 5+5 }{ 2 } \),\(\frac { 4-1 }{ 2 } \)) = (\(\frac { 10 }{ 2 } \),\(\frac { 3 }{ 2 } \)) = (5,\(\frac { 3 }{ 2 } \))
Mid point of PS (D) = (\(\frac { 5-1 }{ 2 } \),\(\frac { -1-1 }{ 2 } \)) = (\(\frac { 4 }{ 2 } \),\(\frac { -2 }{ 2 } \)) = (2,-1)


img 355
AB = BC = CD = AD = \(\sqrt{\frac{61}{4}}\)
Since all the four sides are equal,
∴ ABCD is a rhombus.

Question 2.
The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.
Answer:
Let the vertices A(2,1), B(3, – 2) and C(x, y)
Area of a triangle = 5 sq. unit

\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 5
\(\frac { 1 }{ 2 } \) [-4 + 3y + x – (3 – 2x + 2y)] = 5
-4 + 3y + x – 3 + 2x – 2y = 10
3x + y – 7 = 10
3x + y = 17 ……(1)
Given y = x + 3
Substitute the value ofy = x + 3 in (1)
3x + x + 3 = 17
4x = 17 – 3
4x = 14
x = \(\frac { 14 }{ 4 } \) = \(\frac { 7 }{ 2 } \)
Substitute the value of x in y = x + 3
y = \(\frac { 7 }{ 2 } \) + 3 ⇒ y = \(\frac { 7+6 }{ 2 } \) = \(\frac { 13 }{ 2 } \)
∴ The coordinates of the third vertex is (\(\frac { 7 }{ 2 } \),\(\frac { 13 }{ 2 } \))

Question 3.
Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0
Answer:
3x + y = 2 ……..(1)
5x + 2y = 3 ………(2)
2x – y = 3 ……….(3)
Solve (1) and (2) to get the vertices B


Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = – 1
The point B is (1, – 1)
Solve (2) and (3) to get the vertices C

Substitute the value of x = 1 in (3)
2(1) – y = 3 ⇒ -y = 3 – 2
– y = 1 ⇒ y = – 1
The point C is (1, – 1)
Solve (1) and (3) to get the vertices A

Substitute the value of x = 1 in (1)
3(1) + y = 2
y = 2 – 3 = -1
The point A is (1, – 1)
The points A (1, – 1), B (1, -1), C(1, -1)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of the triangle = 0 sq. units.
Note: All the three vertices are equal, all the point lies in a same points.

Question 4.
If vertices of a quadrilateral are at A(- 5, 7), B(- 4, k), C(- 1, – 6) and D(4, 5) and its area is 72 sq.units. Find the value of k.
Answer:

Area of the quadrilateral ABCD = 72 sq. units.
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 72

-5k + 24 – 5 + 28 – (- 28 – K – 24 – 25) = 144
– 5k + 47 – k – 77 = 144
– 5k + 47 + k + 77 = 144
– 4k + 124 = 144
-4k = 144 – 124
– 4k = 20
k = -5
The value of k = – 5

Question 5.
Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
Answer:
The vertices A(-2, -1), B(4, 0), C(3, 3) and D(- 3, 2)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 0+1 }{ 4+2 } \) = \(\frac { 1 }{ 6 } \)
Slope of BC = \(\frac { 3-0 }{ 3-4 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of CD = \(\frac { 2-3 }{ -3-3 } \) = \(\frac { -1 }{ -6 } \) = \(\frac { 1 }{ 6 } \)
Slope of AD = \(\frac { 2+1 }{ -3+2 } \) = \(\frac { 3 }{ -1 } \) = -3
Slope of AB = Slope of CD = \(\frac { 1 }{ 6 } \)
∴ AB || CD ……(1)
Slope of BC = Slope of AD = -3
∴ BC || AD …..(2)
From (1) and (2) we get ABCD is a parallelogram.

Question 6.
Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Answer:
Let the “x” intercept be “a”
y intercept = 1 – a (sum of the intercept is 1)
Product of the intercept = – 6
a (1 – a) = – 6 ⇒ a – a² = – 6
– a² + a + 6 = 0 ⇒ a² – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a – 3 = 0 (or) a + 2 = 0
a = 3 (or) a = -2
When a = 3
x – intercept = 3
y – intercept = 1 – 3 = – 2
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 3 } \) + \(\frac { y }{ -2 } \) = 1
\(\frac { x }{ 3 } \) – \(\frac { y }{ 2 } \) = 1
2x – 3y = 6
2x – 3y – 6 = 0

When a =-2
x – intercept = -2
y – intercept = 1 – (- 2) = 1 + 2 = 3
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ -2 } \) + \(\frac { y }{ 3 } \) = 1
– \(\frac { x }{ 2 } \) + \(\frac { y }{ 3 } \) = 1
– 3x + 2y = 6
3x – 2y + 6 = 0

Question 7.
The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹ 14/litre and 1220 litres of milk each week at ₹ 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹ 17/litre?
Answer:
Let the selling price of a milk be “x”
Let the demand be “y”
We have to find the linear equation connecting them
Two points on the line are (14, 980) and (16,1220)
Slope of the line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 1220-980 }{ 16-14 } \) = \(\frac { 240 }{ 2 } \) = 120
Equation of the line is y – y₁ = m (x – x₁)
y – 980 = 120 (x – 14) ⇒ y – 980 = 120 x – 1680
-120 x + y = -1680 + 980 ⇒ -120 x + y = -700 ⇒ 120 x – y = 700
Given the value of x = 17
120(17) – y = 700
-y = 700 – 2040 ⇒ – y = – 1340
y = 1340
The demand is 1340 liters

Question 8.
Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer:
Let the image of P(3, 8) and P’ (a, b)
Let the point of intersection be O

Slope of x + 3y = 7 is – \(\frac { 1 }{ 3 } \)
Slope of PP’ = 3 (perpendicular)
Equation of PP’ is
y – y₁ = m(x – x₁)
y – 8 = 3 (x – 3)
y – 8 = 3x – 9
-8 + 9 = 3x – y
∴ 3x – y = 1 ………(1)
The two line meet at 0

Substitute the value of x = 1 in (1)
3 – y = 1
3 – 1 = y
2 = y
The point O is (1,2)
Mid point of pp’ = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(1,2) = (\(\frac { 3+a }{ 2 } \),\(\frac { 8+b }{ 2 } \))
∴ \(\frac { 3+a }{ 2 } \) = 1 ⇒ 3 + a = 2
a = 2 – 3 = -1
\(\frac { 8+b }{ 2 } \) = 2
8 + b = 4
b = 4 – 8 = – 4
The point P’ is (-1, -4)

Question 9.
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer:
Given lines

Substitute the value of y = \(\frac { 5 }{ 13 } \) in (2)
2x – 3 × \(\frac { 5 }{ 13 } \) = -1
2x – \(\frac { 15 }{ 13 } \) = -1
26x – 15 = -13
26x = -13 + 15
26x = 2
x = \(\frac { 2 }{ 26 } \) = \(\frac { 1 }{ 13 } \)
The point of intersection is (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))

Let the x – intercept and y intercept be “a”
Equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (equal intercepts)
It passes through (\(\frac { 1 }{ 13 } \),\(\frac { 5 }{ 13 } \))
\(\frac { 1 }{ 13a } \) + \(\frac { 5 }{ 13a } \) = 1
\(\frac { 1+5 }{ 13a } \) = 1
13a = 6
a = \(\frac { 6 }{ 13 } \)
The equation of the line is

Question 10.
A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.
Answer:
Two straight path will intersect at one point.
Solving this equations

2x – 3y + 4 = 0

Substitute the value of x = \(\frac { -1 }{ 17 } \) in (2)

The point of intersection is (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
Any equation perpendicular to 6x – 7y + 8 = 0 is 7x + 6y + k = 0
It passes through (-\(\frac { 1 }{ 17 } \),\(\frac { 22 }{ 17 } \))
7(-\(\frac { 1 }{ 17 } \)) + 6 (\(\frac { 22 }{ 17 } \)) + k = 0
Multiply by 17
-7 + 6 (22) + 17k = 0
-7 + 132 + 17k = 0
17k = -125 ⇒ k = – \(\frac { 125 }{ 17 } \)
The equation of a line is 7x + 6y – \(\frac { 125 }{ 17 } \) = 0
119x + 102y – 125 = 0
∴ Equation of the path is 119x + 102y – 125 = 0

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions

Question 1.
If the three points (-3, 7), (a, 1), (-3, 2) are collinear then the value of “a” is
(1) 0
(2) -1
(3) -3
(4) 1
Answer:
(3) -3
Hint:
Since the three points are collinear
Area of a ∆ = 0

-3 + 2a – 21 – (7a – 3 – 6) = 0 ⇒ 2a – 24 – 7a + 9 = 0
– 5a – 15 = 0 ⇒ – 5(a + 3) = 0
a + 3 = 0 ⇒ a = -3

Question 2.
If A (5, 5), B (-5, 1), C (10, 7) lie in a straight line, then the area of ∆ ABC is …………….
(1) \(\frac { 13 }{ 2 } \) sq.units
(2) 9 sq.units
(3) 25 sq.units
(4) 0
Answer:
(4) 0
Hint:
Area of the ∆^le

Question 3.
In a rectangle ABCD, area of ∆ ABC is \(\frac { 31 }{ 2 } \) sq. units. Then the area of rectangle is ……………
(1) 62 sq. units
(2) 31 sq. units
(3) 60 sq. units
(4) 30 sq. units
Answer:
(2) 31 sq. units
Hint:
In a rectangle area of ∆ ABC and area of ∆ ACD are equal.
Area of rectangle ABCD = 2 × \(\frac { 31 }{ 2 } \) = 31 sq.units

Question 4.
If the points (k, 2k), (3k, 3k) and (3,1) are collinear, then k is ……………..
(1) \(\frac { 1 }{ 3 } \)
(2) – \(\frac { 1 }{ 3 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) – \(\frac { 2 }{ 3 } \)
Answer:
(2) – \(\frac { 1 }{ 3 } \)
Hint:
Since the three points are collinear. Area of a ∆ = 0

3k² + 3k + 6k – (6k² + 9k + k) = 0 ⇒ 3k² + 9k – 6k² – 10k = 0
-3 k² – k = 0 ⇒ -k(3k + 1) = 0
3k + 1 = 0 ⇒ 3 k = -1 ⇒ k = – \(\frac { 1 }{ 3 } \)

Question 5.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units then x = ………….
(1) 2
(2) \(\frac { 3 }{ 5 } \)
(3) 3
(4) 5
Answer:
(1) 2
Hint:
Area of the triangle = 5 sq. units

6x – 2 + 6x – (-4x + 18 + x) = 10 ⇒ 12x – 2 – (-3x + 18) = 10
12x – 2 + 3x – 18 = 10
15x – 20 = 10 ⇒ 15x = 10 + 20 = 30
x = \(\frac { 30 }{ 15 } \) = 2

Question 6.
The slope of a line parallel to y-axis is equal to …………..
(1) 0
(2) -1
(3) 1
(4) not defined
Answer:
(4) not defined

Question 7.
In a rectangle PQRS, the slope of PQ = \(\frac { 5 }{ 6 } \) then the slope of RS is ………..
(1) \(\frac { -5 }{ 6 } \)
(2) \(\frac { 6 }{ 5 } \)
(3) \(\frac { -6 }{ 5 } \)
(4) \(\frac { 5 }{ 6 } \)
Answer:
\(\frac { 5 }{ 6 } \)
Hint:
In a rectangle opposite sides are parallel.
∴ Slope of the line RS is \(\frac { 5 }{ 6 } \).

Question 8.
The y – intercept of the line y = 2x is ………
(1) 1
(2) 2
(3) \(\frac { 1 }{ 2 } \)
(4) 0
Answer:
(4) 0

Question 9.
The straight line given by the equation y = 5 is …………..
(1) Parallel to x – axis
(2) Parallel to y – axis
(3) Passes through the origin
(4) None of these
Answer:
(1) Parallel to x – axis

Question 10.
The x – intercept of the line 2x – 3y + 5 = 0 is ………….
(1) \(\frac { 5 }{ 2 } \)
(2) \(\frac { -5 }{ 2 } \)
(3) \(\frac { 2 }{ 5 } \)
(4) \(\frac { -2 }{ 5 } \)
Answer:
(2) \(\frac { -5 }{ 2 } \)
Hint:
2x – 3y + 5 = 0 ⇒ 2x – 3y = – 5

Question 11.
The lines 3x – 5y + 1 = 0 and 5x + ky + 2 = 0 are perpendicular if the value of k is ………..
(1) -5
(2) 3
(3) -3
(4) 5
Answer:
(2) 3
Hint:
Slope of the first line (m₁) = \(\frac { -3 }{ -5 } \) = \(\frac { 3 }{ 5 } \)
Slope of the second line (m₂) = \(\frac { -5 }{ k } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 3 }{ 5 } \) × \(\frac { -5 }{ k } \) = -1 ⇒ \(\frac { -3 }{ k } \) = -1
-k = -3 ⇒ The value of k = 3

Question 12.
If x – y = 3 and x + 2y = 6 are the diameters of a circle then the centre is at the point ………..
(1) (0, 0)
(2) (1, 2)
(3) (1, -1)
(4) (4, 1)
Answer:
(4) (4, 1)
Hint:
Centre of the circle is the intersection of the two diameters.

Centre of the circle is (4, 1)

Question 13.
The line 4x + 3y – 12 = 0 meets the x-axis at the point ……….
(1) (4, 0)
(2) (3, 0)
(3) (-3, 0)
Answer:
(2) (3,0)
Hint:
4x + 3y – 12 = 0 meet the x-axis the value of y = 0
4x- 12 = 0 ⇒ 4x = 12
x = \(\frac { 12 }{ 4 } \) = 3 ⇒ The point is (3, 0)

Question 14.
The equation of a straight line passing through the point (2, -7) and parallel to x-axis is ……………….
(1) x = 2
(2) x = -7
(3) y = -7
(4) y = 2
Answer:
(3) y = -7
Hint:
Equation of a line parallel to x-axis is y = -7

Question 15.
The equation of a straight line having slope 3 and y intercept – 4 is ………………
(1) 3x – y – 4 = 0
(2) 3x + y – 4 = 0
(3) 3x – y + 4 = 0
(4) 3x – y + 4 = 0
Answer:
(1) 3x – y – 4 = 0
Hint. The equation of a line is y = mx + c
y = 3 (x) + (-4) ⇒ y = 3x – 4
3x – y – 4 = 0

II. Answer the following questions:

Question 1.
If the points (3, – 4) (1, 6) and (- 2, 3) are the vertices of a triangle, find its area.
Answer:
Let the vertices A (3, – 4), B (1, 6) and C (- 2, 3)

Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁, – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of a ∆ = 18 sq. units

Question 2.
If the area of the triangle formed by the points (1,2) (2,3) and (a, 4) is 8 sq. units, find a.
Answer:
Area of a triangle = 8 sq. units.

\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 8.
\(\frac { 1 }{ 2 } \) [3 + 8 + 2a – (4 + 3a + 4)] = 8
11 + 2a – 8 – 3a= 16 ⇒ – a + 3 = 16
– a = 16 – 3 ⇒ a = -13
The value of a = -13

Question 3.
If the points A (2, 5), B (4, 6) and C (8, a) are collinear find the value of “a” using slope concept.
Answer:
Since the three points are collineal
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = Slope of BC
\(\frac { 6-5 }{ 4-2 } \) = \(\frac { a-6 }{ 8-4 } \) ⇒ \(\frac { 1 }{ 2 } \) = \(\frac { a-6 }{ 4 } \) ⇒ 2a – 12 = 4 ⇒ 2a = 16
a = \(\frac { 16 }{ 2 } \) = 8 ⇒ The value of a = 8

Question 4.
If the points (x,y) is collinear with the points (a, 0) and (0, b) then prove that \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
Answer:
Let A (x, y), B (a, 0), C(0, b)
Since the three points are collinear
Slope of AB = Slope of BC
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(\frac { 0-y }{ a-x } \) = \(\frac { b-0 }{ 0-a } \)
\(\frac { -y }{ a-x } \) = \(\frac { b }{ -a } \)
ay = b (a – x)
ay = ba – bx
ay + bx = ab
Divided by ab
\(\frac { ay }{ ab } \) + \(\frac { bx }{ ab } \) = \(\frac { ab }{ ab } \)
\(\frac { y }{ b } \) + \(\frac { x }{ a } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1

Question 5.
A straight line passes through (1, 2) and has the equation y – 2x – k = 0. Find k.
Answer:
The given line is y – 2x – k = 0
It passes through (1,2)
(2) -2 (1) -k = 0 ⇒ 2 – 2 – k = 0
0 – k = 0 ⇒ k = 0
The value of k = 0

Question 6.
If a line passes through the mid point of AB where A is (3, 0) and B is (5, 4) and makes an angle 60° with x – axis find its equation.
Answer:

Slope of a line = tan 60°
= \(\sqrt { 3 }\)
Equation of a line is y – y₁ = m (x – x₁)
y – 2 = \(\sqrt { 3 }\) (x – 4)
y – 2 = \(\sqrt { 3 }\) x – 4 \(\sqrt { 3 }\)
\(\sqrt { 3x }\) – y + 2 – 4\(\sqrt { 3 }\) = 0

Question 7.
Find the equation of the line through (3, 2) and perpendicular to the line joining (4, 5) and (1,2)
Answer:
Slope of a line = \(\frac { 2-5 }{ 1-4 } \) ⇒ \(\frac { -3 }{ -3 } \) = 1
Slope of the line perpendicular to it is – 1
Equation of the line joining -1 and (3, 2) is
y – y₁ = m (x – x₁) ⇒ y – 2 = -1(x – 3)
y – 2 = -x + 3 ⇒ x + y – 5 = 0

Question 8.
P and Q trisect the line segment joining the points (2, 1) and (5, – 8). If the point P lies on 2x – y + k = 0, then find the value of k.
Answer:

A line divides internally in the ratio 1 : 2
A line divide internally in the ratio l : m

The point P = (\(\frac { 5+4 }{ 3 } \),\(\frac { -8+2 }{ 3 } \))
= (\(\frac { 9 }{ 3 } \),\(\frac { -6 }{ 3 } \)) = (3, -2)
The given line 2x – y + k = 0 passes through the point (3,-2)
2 (3) – (- 2) + k = 0
6 + 2 + k = 0
8 + k = 0
k = – 8
The value of k = – 8

Question 9.
The line 4x + 3y – 12 = 0 intersect the X, Y – axis at A and B respectively. Fine the area of ∆AOB.
Answer:
The equation of the line AB is 4x + 3y – 12 = 0

4x + 3y = 12
\(\frac { 4x }{ 12 } \) + \(\frac { 3y }{ 12 } \) = 1 ⇒ \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) = 1
The point A is (3, 0) (it intersect the X – axis)
and B is (0, 4) (it intersect the Y – axis)
Area of ∆ AOB = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ (x₂y₁ + x₃y₂ + x₁y₃)]

Question 10.
Find the equation of the line passing through (4, 5) and making equal intercept in the axes.
Answer:
Let the equal intercept on the axes be a, a.
Equation of the line is \(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (Given equal intercepts)
The line passes through (4, 5)
\(\frac { 4 }{ a } \) + \(\frac { 5 }{ a } \) = 1 ⇒ \(\frac { 9 }{ a } \) = 1 ⇒ a = 9
The equation of the line is \(\frac { x }{ 9 } \) + \(\frac { y }{ 9 } \) = 1
Multiply by 9
x + y – 9 = 0

Question 11.
Find the equation of the line passing through (2, – 1) and whose intercepts on the axes are equal in magnitude but opposite in sign.
Answer:
Let the x – intercept be “a” and y intercept be = “-a”
The equation of the line is
\(\frac { x }{ a } \) + \(\frac { y }{ -a } \) = 1 (y – intercept is – a)
\(\frac { x }{ a } \) – \(\frac { y }{ a } \) = 1
It passes through (2, -1)
\(\frac { 2 }{ a } \) – \(\frac { (-1) }{ a } \) = 1
\(\frac { 2 }{ a } \) + \(\frac { 1 }{ a } \) = 1 ⇒ \(\frac { 3 }{ a } \) = 1
a = 3
The equation of the line is
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 3 } \) + \(\frac { y }{ -3 } \) = 1 ⇒ \(\frac { x }{ 3 } \) – \(\frac { y }{ 3 } \) = 1
x – y = 3
The equation is x – y – 3 = 0

Question 12.
The straight line cuts the coordinate axes at A and B. If the mid point of AB is (3,2) then find the equation of AB.
Answer:
Let the point A be (a, 0) and B be (0, b)


The point A (6, 0) and B (0, 4)
Equation of the line AB is

III. Answer the following questions

Question 1.
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the ‘ coordinates of any point “c”, if AC = BC and Area of triangle ABC = 10 sq. units.
Answer:
Let the coordinates C be (a, 6) then AC = BC
AC² = BC²
(a – 3)² + (b – 4)² = (a – 5)² + (b + 2)²
a² + 9 – 6a + b² + 16 – 8b = a² + 25 – 10a + b² + 4 – 4b
a² + b² + 25 – 6a – 86 = a² + b² + 29 – 10a + 4b

25 – 6a – 8b = 29 – 10a + 46
4a – 12b = 4 ⇒ a – 3b = 1 ………… (1)
Area of ∆ ABC = 10 sq. units
\(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)] = 10

-6 + 5b + 4a – (20 – 2a + 3b) = 20
-6 + 5b + 4a – 20 + 2a – 3b = 20
6a + 2b – 26 = 20 ⇒ 6a + 2b = 46

Substitute the value of a = 7 in (2)
3 (7) + b = 23 ⇒ b = 23 – 21 = 2
The coordinate C is (7, 2)

Question 2.
The four vertices of a Quadrilateral are (1,2) (- 5,6) (7, – 4) and (k, – 2) taken in order. If the area of the Quadrilateral is 9 sq. units, find the value of k.
Answer:
Let A (1, 2) B (- 5, 6) C (7, – 4) and D (k, – 2)
Area of the
Quadrilateral ABCD = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]

Area of the Quadrilateral ABCD = 3k – 9
Given area of a Quadrilateral is 9 sq. units.
3k – 9 = 9 ⇒ 3k = 18 ⇒ k = \(\frac { 18 }{ 3 } \) = 6
The value of k = 6

Question 3.
Find the area of a triangles whose three sides are having the equations x + y = 2, x – y = 0 and x + 2y – 6 = 0.
Answer:
Find the three vertices of the triangles by solving their equation.
To find vertices A


Substitute the value of y = 4 in (1)
x + 4 = 2 ⇒ x = 2 – 4 = -2
The vertices A is (- 2, 4)
To find vertices B

Substitute the value of x = 1 in (1)
1 + y = 2 ⇒ y = 2 – 1 = 1
The vertices B is (1, 1)
To find vertices C

y = \(\frac { 6 }{ 3 } \) = 2
Substitute the value y = 2 in (3)
x – 2 = 0 ⇒ x = 2
The vertices C is (2, 2)

Area of the ∆ BC = 3 sq. units

Question 4.
Verify the Median of a triangle divides into two triangles of equal areas whose vertices are A (4, – 6), B (3, – 2) and C (5, 2)
Answer:

Let D be the mid point of AC .
Mid point of AC = (\(\frac { 5+4 }{ 2 } \),\(\frac { 2-6 }{ 2 } \)) = (\(\frac { 9 }{ 2 } \),-2)
Area of the triangle = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of ∆ ADB = Area of ∆ BDC
A median divides the triangle of equal areas.

Question 5.
Find the area of the ∆ ABC with A (1, – 4) and the mid points of sides through A being (2,-1) and (0,-1)
Answer:
Let the coordinates of B and C are (a, b) and (c, d) respectively.
Sides through A are AB and AC

Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
(2, -1) = (\(\frac { 1+a }{ 2 } \),\(\frac { -4+b }{ 2 } \))
\(\frac { 1+a }{ 2 } \) = 2
1 + a = 4
a = 4 – 1
= 3
The point B is (3,2)
\(\frac { -4+b }{ 2 } \) = -1
-4 + b = -2
b = -2 + 4
= 2
Mid point of AC = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))
(0,-1) = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))
\(\frac { 1+c }{ 2 } \) = 0
1 + c = 0
c = 0 – 1
= – 1
The point C is (-1,2)
\(\frac { -4+d }{ 2 } \) = -1
– 4 + d = -2
d = – 2 + 4
= 2
Thus the coordinates of the vertices of ∆ ABC are A (1, – 4) B (3, 2) and C (- 1, 2)
Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x₁y₂ + x₂y₃ + x₃y₁ – (x₂y₁ + x₃y₂ + x₁y₃)]

Area of ∆ ABC = 12 sq. units

Question 6.
Find the equation of the straight lines passing through (- 3, 10) whose sum of the intercepts is 8.
Answer:
Let the “x” intercept be “a” and y intercept be “b”
Sum of the intercepts = 8
a + b = 8 ⇒ b = 8 – a
Equation of a line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ 8-a } \) = 1
It passes through (-3,10)
\(\frac { -3 }{ a } \) + \(\frac { 10 }{ 8-a } \) = 1
\(\frac { -3(8-a)+10a }{ a(8-a) } \) = 1
-24 + 3a + 10a = 8a – a²
-24 + 13a = 8a – a²
a² + 5a – 24 = 0 ⇒ (a + 8) (a – 3) = 0
a + 8 = 0 (or) a – 3 = 0 ⇒ a = -8 (or) a = 3
The equation of a line is a
a = -8
\(\frac { x }{ -8 } \) + \(\frac { y }{ 8+8 } \) = 1
\(\frac { x }{ -8 } \) + \(\frac { y }{ 16 } \) = 1
-2x + y = 16
2x – y + 16 = 0
a = 3
\(\frac { x }{ 3 } \) + \(\frac { y }{ 5 } \) = 1
5x + 3y = 15
5x + 3y – 15 = 0
The equation of the lines are 2x – y + 16 = 0 (or) 5x + 3y – 15 = 0.

Question 7.
If (5, – 3), (- 5, 3), (6, 6) are the mid points of the sides of a triangle, find the equation of the sides.
Answer:
Since D, E, F are the mid points of ∆ ABC
EF || AB, FD || BC and DE || AC
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of EF = \(\frac { 6-3 }{ 6+5 } \) = \(\frac { 3 }{ 11 } \)
Since EF || AB; Slope of AB = \(\frac { 3 }{ 11 } \)
Equation of AB is
y – y₁ = m (x – x₁)
y + 3 = \(\frac { 3 }{ 11 } \) (x – 5)
3x – 15 = 11y + 33
3x – 11y – 15 – 33 = 0
3x – 11y – 48 = 0
Slope of DE = Slope of AC
Slope of DE = \(\frac { 3+3 }{ -5-5 } \) = \(\frac { 6 }{ -10 } \) = –\(\frac { 6 }{ 10 } \) = –\(\frac { 3 }{ 5 } \)
Slope of AC = – \(\frac { 3 }{ 5 } \)
Equation of AC is
y – y₁ = m (x – x₁)

y – 6 = – \(\frac { 3 }{ 5 } \) (x – 6) ⇒ 5y – 30 = -3x + 18
3x + 5y – 30 – 18 = 0 ⇒ 3x + 5y – 48 = 0
Slope of DF = Slope of BC
Slope of DF = \(\frac { 6+3 }{ 6-5 } \) = \(\frac { 9 }{ 1 } \) = 9
Slope of BC = 9
Equation of the line BC is
y – y₁ = m(x – x₁)
y – 3 = 9 (x + 5) ⇒ 9x + 45 = y – 3
9x – y + 45 + 3 = 0 ⇒ 9x – y + 48 = 0
Equation of the sides are
3x – 11y – 48 = 0 ; 9x – y + 48 = 0 and 3x + 5y – 48 = 0

Question 8.
Find the equation of the straight line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and is perpendicular to the line joining the points (5,1) and (-2, 2)
Answer:

Substitute the value of x in (1)
5 (5) – 8y = – 23 ⇒ 25 – 8y = – 23
-8y = – 23 – 25 ⇒ -8y = – 48
y = \(\frac { 48 }{ 8 } \) = 6
The point of intersection is (5,6)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of the line joining the points (5,1) and (-2,2) = \(\frac { 2-1 }{ -2-5 } \)
= \(\frac { 1 }{ -7 } \) = – \(\frac { 1 }{ 7 } \)
Slope of the perpendicular line is = 7
Equation of a line is
y – y₁ = m(x – x₁) ⇒ y – 6 = 7 (x – 5)
y – 6 = 7x – 35 ⇒ -7x + y – 6 + 35 = 0
7x – y – 29 = 0

Question 9.
Find the equation of the line passing through the point of intersection of 4x – y – 3 = 0 and x + y – 2 = 0 and perpendicular to 2x – 5y + 3 = 0.
Answer:

x = \(\frac { 5 }{ 5 } \) = 1
Substitute the value of x = 1 in (2)
1 + y = 2
y = 2 – 1 = 1
The point of intersection is (1, 1)
Any line perpendicular to 2x – 5y + 3 = 0 is
5x + 2y + k = 0
It passes through (1,1)
5(1) + 2(1) + k = 0 ⇒ 5 + 2 + k = 0
7 + k = 0 ⇒ k = -7
The line is 5x + 2y – 7 = 0

Question 10.
Find the equation of the line through the point of intersection of the lines 2x + y – 5 = 0 and x + y – 3 = 0 and bisecting the line segment joining the points (3, – 2) and (- 5, 6).
Answer:

x = 2
Substitute the value of x = 2 in (2)
2 + y = 3
y = 3 – 2 = 1
The point of intersection is (2, 1)
Mid point of the line joining the points (3,-2) and (-5,6)

Mid point of the line
Equation of the line joining the points (2, 1) and (-1,2) is

x – 2 = -3 (y – 1)
x – 2 = -3y + 3
x + 3y – 5 = 0
The equation of the line is x + 3y – 5 = 0