Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.10 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.10

Question 1.

Determine whether the following measurements produce one triangle, two triangles or no triangle. ∠B = 88°, a = 23 , b = 2. Solve if solution exists.

Answer:

Using sine formula

= 23 × 0.999

= 22.99

which is not possible

∴ Solution of the given triangle does not exsit.

Question 2.

If the sides of a ∆ ABC are a = 4, b = 6 and C = 8, then show that 4 cos B + 3 cos C = 2.

Answer:

In ∆ ABC Given that a = 4, b = 6, c = 8

Using cosine formula

Question 3.

In a ∆ ABC, if a = √3 – 1, b = √3 + 1 and C = 60° find the other side and other two angles.

Answer:

In a ∆ ABC, Given

a = √3 – 1, b = √3 + 1

C = 60°

Using cosine formula a

C^{2} = a^{2} + b^{2} – 2 ab cos C

= (√3 – 1)^{2} + (√3 + 1)^{2} – 2(√3 – 1) × (√3 + 1) cos 6o°

= 3 – 2√3 + 1 + 3 + 2√3 + 1 – 2 (3 – 1) × \(\frac { 1 }{ 2 }\)

c^{2} = 8 – 2 = 6 ⇒ c = √6

Using sine formula

sin (45° – 30°) = sin 45° . cos 30° – cos 45° sin 30°

From equations (1) and (2), we have

sin A = sin 15° ⇒ A = 15°

In ∆ ABC, we have A + B + C = 180°

15° + B + 60° = 180°

B = 180°- 75°

B = 105°

∴ The required sides and angles are

c = √6, A = 15°, B = 105°

Question 4.

In any ∆ ABC, prove that the area

Answer:

Area of ∆ ABC is ∆ = \(\frac { 1 }{ 2 }\) bc = sin A

Using cosine formula

Question 5.

In a ∆ABC, if a = 12 cm, b = 8 cm and C = 30°, then show that its area is 24 sq.cm.

Answer:

In ∆ ABC Given

a = 12 cm ,

b = 8 cm,

C = 30°

Question 6.

In a ∆ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.

Answer:

In a ∆ ABC, Given a = 18 cm, b = 24cm and c = 30 cm

Area of the triangle ABC

Question 7.

Two soldiers A and B in two different underground bunkers on a straighi road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5km apart, find the distance of the intruder from B.

Answer:

Let A and B be the two positions of the soldiers.

AC – direction of the intruder seen from A.

BC – the direction of the intruder seen from B.

∠ BAC = 30° angle of elevation of the intruder from A.

∠ PBC = 45° angle of elevation of the intruder from B.

Distance between A and B = 5k.m.

In ∆ ABC, ∠ ABC = 180° – 45° = 135°

∠ BCA = 180° – ( 135° + 30°)

= 180° – 165° = 15°

Using sine formula

Question 8.

A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be 8 km, while the distance to the westernmost point from P to be 6 km. If the angle between the two lines of sight is 60°, find the width of the pond.

Answer:

A – be the easternmost point on the pond and

B – be the westernmost point on the pond.

AB – Width of the pond

P – Point of observation.

The distance of A from P = 8 km

Distance of B from P = 6km

Angle between the directions PA and PB

∠APB = 60°

In ∆ PAB, using cosine formula

AB^{2} = PA^{2} + PB^{2} – 2PA . PB . cos ∠APB

AB^{2} = 8^{2} + 6^{2} – 2 × 8 × 6 . cos 60°

= 64 + 36 – 96 × \(\frac { 1 }{ 2 }\)

= 100 – 48 = 52

AB = \(\sqrt{52}\) = \(\sqrt{4 \times 13}\)

AB = 2\(\sqrt{13}\) k.m.

Width of the pond = 2\(\sqrt{13}\) k.m

Question 9.

Two Navy helicopters A and B are flying over the Bay of Bengal at saine altitude from sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B.

Answer:

A , B are the positions of the helicopter above the sea level.

Distance between A and B = 10 km

C – Position of the boat on the surface of sea.

AC, BC are the directions of the boat as seen from A and B respectively.

Distance of the boat C from A = 6 k.m

∠ ACB = 60°

Using cosine formula

AB^{2} = BC^{2} + AC^{2} – 2 BC . AC cos ∠ACB

c^{2} = a^{2} + b^{2} – 2 ab cos C

10^{2} = a^{2} + 6^{2} – 2a × 6 cos 60°

100 = a^{2} + 36 – 12a\(\left(\frac{1}{2}\right)\)

0 = a^{2} + 36 – 6a – 100

a^{2} – 6a – 64 = 0

Question 10.

A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If AP = 3 km, BP = 5 km, and ∠APB = 120°, then find the length of the tunnel to be built.

Answer:

p^{2} = a^{2} + b^{2} – 2ab cos P

p^{2} = 9 + 25 – 30 Cos 120°

p^{2} = 9 + 25 – 30 (-1/2) = 34 + 15 = 49

⇒ p = \(\sqrt{49}\) = 7 km

Question 11.

A farmer wants to purchase a triangular-shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60°. If the land costs Rs.500 per square feet, find the amount he needed to purchase the land. Also, find the perimeter of the land.

Answer:

Let ∆ ABC be the shape of the land.

Given AB = 120 ft, AC = 60ft

∠ BAC = 60°

Using cosine formula in ∆ ABC

BC^{2} = AB^{2} + AC^{2} – 2AB . AC cos ¿BAC

BC^{2} = 120^{2} + 60^{2} – 2 × 120 × 60 cos (60°)

= 14400 + 3600 – 14400 × \(\frac { 1 }{ 2 }\)

= 18000 – 7200

BC^{2} = 10800 = 100 × 2 × 2 × 3 × 3 × 3

BC^{2} = 10^{2} × 2^{2} × 3^{2} × 3

BC = \(\sqrt{10^{2} \times 2^{2} \times 3^{2} \times 3}\)

BC = 10 × 2 × 3√3

BC = 60√3 k.m.

Perimeter of the Land = AB + BC + AC

= 120 + 60√3 + 60

= 180 + 60√3

= 60 (3 + √3) feet.

Area of ∆ ABC = \(\frac { 1 }{ 2 }\) × AB × AC × sin ∠ BAC

= \(\frac { 1 }{ 2 }\) × 60 × 120 sin 60°

= 30 × 120 × \(\frac{\sqrt{3}}{2}\)

= 30 × 60 × √3

= 1800 √3 sq. feet.

Cost of 1 sq. feet Rs. 500

∴ Cost of 800 √3 sq. feet = 800 √3 × 500 = 900000√3

Total amount needed = Rs. 900000√3

Perimeter of the land = 60(3 + √3)feet.

Question 12.

A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30°. If after 100 km, the target has an angle of depression of 45°, how far is the target from the fighter jet at that instant?

Answer:

Let A be the position of the jet fighter observing the target at an angle of depression 30°.

Also, Let B be the position of the jet 100 k.m away horizontally from A observing the target at an angle of depression 45°.

In ∆ TAB, AB = 100 k.m

∠TAB = 30°

∠ABT = 180°- 45° = 135°

∠ATB = 180° – (135°+ 300) = 180° – 165° = 15°

Question 13.

A plane is 1 km from one landmark and 2 km from another. From the plane’s point of view, the land between them subtends an angle of 45°. How far apart are the landmarks?

Answer:

A, B are the two landmarks,

C – Position of the plane.

The distance of the plane from the landmark A = 1 k.m

The distance of the plane from the landmark B = 2 k.m

∠ACB = 45°

From the ∆ ABC, using cosine formula

AB^{2} = AC^{2} + BC^{2} – 2AC. BC. cos45°

= 1^{2} + 2^{2} – 2 × 1 × 2

AB^{2} = 1 + 4 – 2 × √2 = 5 – 2√2

AB = \(\sqrt{5-2 \sqrt{2}}\)

Distance between the landmarks AB = \(\sqrt{5-2 \sqrt{2}}\) km.

Question 14.

A man starts his morning walk at a point A reaches two points B and C and finally back to A such that ∠A = 60° and ∠B = 45°, AC = 4km in the ∆ ABC. Find the total distance he covered during his morning walk.

Answer:

Given In ∆ABC

AC = 4 k.m

∠A = 60°,

∠B = 45°

∠C = 180° – (60° + 45°)

∴ ∠C = 180° – 105° = 75°

Using sine formula

Question 15.

Two vehicles leave the same place P at the same time moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour the vehicle reaches destinations A and B. If AB subtends 60° at the initial point P, then find AB.

Answer:

P – Initial point.

PA – The direction of the first vehicle travels with speed km/hr.

PB – The direction of the second vehicle travels with a speed of 80km/hr.

Given in half an hour first vehicle reaches destination A.

∴ PA = \(\frac{60}{2}\) = 30 km.

Also in half an hour the second vehicle reaches the destination B.

∴ PA = \(\frac{80}{2}\) = km.

In ∆ PAB, PA = 30, PB = 40, ∠APB = 60°

Using cosine formula

AB^{2} = PA^{2} + PB^{2} – 2PA PB cos ∠APB

AB = 30^{2} + 40^{2} – 2 × 30 × 40 cos 60°

= 900 + 1600 – 2400 × \(\frac { 1 }{ 2 }\)

= 2500 – 1200

AB^{2} = 1300

AB = \(\sqrt{1300}\) = \(\sqrt{13 \times 100}\)

AB = 10√13 k.m.

Question 16.

Suppose that a satellite in space, an earth station, and the centre of earth all lie in the same plane. Let r be the radius of earth and R he the distance from the centre of earth to the satellite. Let d be the distance from the earth station to the satellite. Let 30° be the angle of elevation from the earth station to the satellite, If the line segment connecting the earth station and satellite subtends angle α at the centre of earth then prove that

Answer:

O – Centre of Earth,

A – Position of Earth station.

S – Position of the satellite.

Given the radius of Earth

OA = r

The angle of elevation of the satellite from the Earth station = 30°

The distance of the satellite from the Earth station AS = d

The distance of the satellite from the centre of the Earth OS = R.

Angle subtended by the line segment AS at the centre of earth ∠AOS = α

In △ AOS, OA = r, AS = d, OS = R, ∠AOS = α

Using cosine formula

AS^{2} = OA^{2} + OS^{2} – 2 OA . OS cos ∠AOS

d^{2} = r^{2} + R^{2} – 2(r) (R) cos α