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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.5

Question 1.
Solve the following systems of linear equations by Gaussian elimination method:
(i) 2x – 2y + 3z = 2, x + 2y – z = 3, 3x – y + 2z = 1
Solution:
Augmented matrix

Writing the equivalent equations from echelon from.
x – y + 2z = 3 …………. (1)
5y – 6z = -4 ………….. (2)
-z = -4
z = 4
(2) ⇒ 5y – 6z = -4
5y – 24 = -4
5y = -4 + 24
5y = 20
y = 4
(1) ⇒ x – y + 2z = 3
x – 4 + 8 = 3
x = 3 + 4 – 8
x = -1
∴ x = -1, y = 4, z = 4

(ii) 2x + 4y + 6z = 22, 3x + 8y + 5z = 27, -x + y + 2z = 2.
Solution:
Augmented matrix

Writing the equivalent equations from echelon from.
x + 2y + 3z = 11 …………. (1)
y – 2z = -3 ………….. (2)
11z = 22
z = 2
(2) ⇒ y – 2z = -3
y – 4 = -3
y = -3 + 4
y = 1
(1) ⇒ x + 2y + 3z = 11
x + 2(1) + 3(2) = 11
x + 2 + 6 = 11
x = 11 – 8 = 3
∴ x = 3, y = 1, z = 2

Question 2.
If ax² + bx + c is divided by x + 3, x – 5, and x – 1, the remainders are 21, 61 and 9 respectively. Find a, b and c. (Use Gaussian elimination method.)
Solution:
Given: f(x) = ax² + bx + c
In Remainder Theorem
f(-3) = 21
a(-3)² + b(-3) + c = 21
9a – 3b + c = 21 ……….. (1)
f(5) = 61
25a + 5b + c = 61 …………. (2)
f(1) = 9
a + b + c = 9 ………… (3)
Augmented matrix

Writing the equivalent equations from echelon from.
a + b + c = 9 …………. (1)
b + 2c = 5 ………….. (2)
-4c = -8
c = 2
(2) ⇒ b + 2c = 5
b + 4 = 5
b = 5 – 4
b = 1
(1) ⇒ a + b + c = 9
a + 1 + 2 = 9
a = 9 – 3
a = 6
a = 6, b = 1, c = 2

Question 3.
An amount of Rs 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is Rs 5,000. The income from the third bond is Rs 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
Solution:
Let the amounts of 3 bounds be x, y, z
x + y + z = 65,000

Writing the equivalent equations from echelon from.
x + y + z = 65000 …………. (1)
2y + 3z = 90000 ………….. (2)
21z = 42000
z = 20000
(2) ⇒ 2y = 90000 – 3 × 20000
2y = 30000
y = 15000
(1) ⇒ x + 15000 + 20000 = 65000
x = 30000
∴ x = 30000, y = 15000, z = 20000

Question 4.
A boy is walking along the path y = ax² + bx + c through the points (-6, 8),(-2, -12), and (3, 8). He wants to meet his friend at P(7, 60). Will he meet his friend? (Use Gaussian elimination method.)
Solution:
y = ax² + bx + c
At(-6, 8) ⇒ 8 = 36a – 6b + c ………… (1)
At(-2, -12) ⇒ -12 = 4a – 2b + c ………… (2)
At(3, 8) ⇒ 8 = 9a + 3b + c ………… (3)
Augmented matrix

Writing the equivalent equations from the echelon.
36a – 6b + c = 8 …………. (1)
3b – 2c = 29 ………….. (2)
5c = -50
c = -10
(2) ⇒ 3b – 2c = 29
3b – 20 = 29
3b = 9
b = 3
(1) ⇒ 36a – 18 – 10 = 8
36a = 8 + 18 + 10
36a = 36
a = 1
At P (7, 60), y = ax² + bx + c
60 = 1(7²) + 3(7) – 10
60 = 49 – 21 – 10
60 = 60
He will meet his friend at P (7, 60)

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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 1.
Solve the following system of linear equations by matrix inversion method.
(i) 2x + 5y = -2, x + 2y = -3
Solution:

(ii) 2x – y = 8, 3x + 2y = -2
Solution:

x = 2, y = -4

(iii) 2x + 3y – z = 9, x + y + z = 9, 3x – y – z = -1
Solution:

|A| = 2(-1+1)-3(-1-3)-1(-1-3)
= 0 + 12 + 4 =16 ≠ 0 A^-1 exists.

∴ x = 2, y = 3, z = 4

(iv) x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
Solution:

AX = B
X = A^-1B
A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
6 & -4 & 5 \\
5 & 2 & 2
\end{array}\right]\)
|A| = 1(-8-10)-1(12-25)+1(12+20)
= 18 + 13 +32 = 27
≠ 0
A^-1 Exists

∴ x = 3, y = -2, z = 1

Question 2.

Find the products AB and BA and hence solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2
Solution:

AB = BA = 4I₃

∴ x = 2, y = 1, z = -1

Question 3.
A man is appointed in a job with a monthly salary of a certain amount and a fixed amount of annual increment. If his salary was Rs 19,800 per month at the end of the first month after 3 years of service and Rs 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use the matrix inversion method to solve the problem.)
Solution:
Let the man starting the salary be Rs x and his annual increment be Rs y.
Given x + 3y = 19,800
x + 9y = 23,400
The equation can be written as
\(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{c}
19800 \\
23400
\end{array}\right]\)
AX = B
X = A^-1B
A = \(\left[\begin{array}{ll}
1 & 3 \\
1 & 9
\end{array}\right]\)
To find A^-1
|A| = 9 – 3 = 6 ≠ 0 A^-1 Exists.

Monthly salary = Rs 18000
Annual increment = Rs 1800

Question 4.
4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by Using the matrix inversion method.
Solution:
Let the time taken by one man alone be x days and one woman alone be y days.


∴ One man can do 18 days
One woman can do 36 days

Question 5.
The prices of three commodities A, B, and C are Rs x,y, and z per unit respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B. Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn Rs 15,000, Rs 1,000 and 14,000 respectively. Find the prices per unit of A, B, and C. (Use the matrix inversion method to solve the problem.)
Solution:
Let x, y, z are commodities of A, B, C
2x – 4y + 5z = 15,000 ………… (1)
3x + y – 2z = 1000 ………… (2)
-x + 3y + z = 4000 ………… (3)

|A| = 2(1 + 6)+ 4(3 – 2) + 5(9 + 1)
= 2(7) + 4(1) + 5(10)
= 14 + 4 + 50 = 68
≠ 0
A^-1 Exists.


x = Rs 2000, y = Rs 1000, z = Rs 3000
The prices per unit of A, B, and C are Rs 2000, Rs 1000, Rs 3000

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 1.
Solve the following systems of linear equations by Cramer’s rule:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
Solution:

(ii) \(\frac{3}{x}\) + 2y =12, \(\frac{2}{x}\) + 3y = 13
Solution:
Let \(\frac{1}{x}\) = a
3a + 2b = 12
2a + 3b = 13

(iii) 3x + 3y – z = 1 1, 2x – y + 2z = 9, 4x + 3y + 2z = 25
Solution:
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= 3(-2 – 6) -3 (4 – 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 – 10 = -22 ≠ 0
Δ_x = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= 11 (-2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 – 52 = -44
Δ_y = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) -11(4) -1(14)
= -96 + 44 – 14 = -66
Δ_x = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
= 3(-25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 – 42 + 110
= -88

Solution:

3a – 4b – 2c = 1 ……….. (1)
a + 2b + c = 2 …………… (2)
2a – 5b – 4c = -1 ………….. (3)
Δ = \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| \)
= 3(-8 + 5) + 4 (-4 – 2) – 2(-5 – 4)
= 3(-3) +4(-6) -2(-9)
= -9 – 24 + 18
= -15 ≠ 0
Δ_a = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= 1(-8 + 5) + 4(-8 + 1) – 2(-10 + 2)
= 1(-3) + 4(-7) – 2(-8)
= -3 – 28 + 16
= -15
Δ_b = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
= 3(-8 + 1) – 1(-4 – 2) – 2(-1 – 4)
= 3(-7) -1(-6) – 2(-5)
= – 21 + 6 + 10 = -5
Δ_c = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4)
= 24 – 20 – 9 = -5

Question 2.
In a competitive examination, one mark is awarded for every correct answer while \(\frac{1}{4}\) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer’s rule to solve the problem).
solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x
and the No. of questions answered wrongly be y
Here, x + y = 100 and x – \(\frac { 1 }{ 4 }\) y = 80
(i.e) x + y = 100 and 4x – y = 320

correct questions = 84
wrong questions = 16.

Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramer’s rule to solve the problem).
solution:
Let two solutions x and y
x + y = 10 …….. (1)
0.25 x + (0.50)y = (0.40) ……….. (2)
(2) × 100 ⇒ 25x + 50y = 400
(2) ÷ 5 ⇒ 5x + 10y = 80 …………. (3)

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer’s rule to solve the problems).
solution:
pump A fills \((\frac {1}{x})^{th}\) of the tank in 1 hour.
pump B fills \((\frac {1}{y})^{th}\) of the tank in 1 hour.
Both can filled \((\frac {1}{10})^{th}\) of the tank in 1 hour.

using Cramer’s rule

Pump A takes 15 minutes
Pump B takes 30 minutes

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
solution:
Let the Cost of dosai, Idlies and vadais be x, y, z
2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= 2(4 – 16) – 3(4 – 20) + 2(8 – 10)
= 2(-12) – 3(-16) + 2(-2)
= -24 + 48 – 4
= 20 ≠ 0
Δ_x = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
= 150(4 – 16) – 3(400 – 1000) + 2(800 – 500)
= 150(-12) – 3(-600) + 2(300)
= -1800 + 1800 + 600
= 600
Δ_y = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| \)
= 2(400 – 1000) – 150(4 – 20) + 2(500 – 1000)
= 2(-600) – 150(-16) + 2(-500)
= -1200 + 2400 – 1000
= 200
Δ_z = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| \)
= 2(500 – 800) – 3(500 – 1000) + 150(8 – 10)
= 2(-300) – 3(-500) + 150(-2)
= -600 + 1500 – 300
= 600

x = Rs 30, y = Rs 10, z = Rs 30
There are 3 dosai, 6 idlies and 6 vadais
= 3x + 6y + 6z
= 3(30) + 6(10) + 6 (30)
= 90 + 60 + 180
= Rs. 330
They can eat within the amount.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Find the rank of the following matrices by minor method:

Solution:

(i) A = \(\begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}\)
A is a matrix of order 2 × 2 and p(A) ≤ 2
Second order minor
|A| = \(\begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}\)
= 4 – 4 = 0
∴p(A) ≠ 2
First order minor is non vanishing
p(A) = 1

(ii) A = \(\left[\begin{array}{rr}
-1 & 3 \\
4 & -7 \\
3 & -4
\end{array}\right]\)
A is a matrix of order 3 × 2 and p(A) ≤ 2
Second order minor
\(\begin{bmatrix} -1 & 3 \\ 4 & -7 \end{bmatrix}\)
= 7 – 12 = -5 ≠ 0
∴ p(A) = 2

(iii) A = \(\left[\begin{array}{rrrr}
1 & -2 & -1 & 0 \\
3 & -6 & -3 & 1
\end{array}\right]\)
A is a matrix of order 2 × 4 and p(A) ≤ 2
Second order minor

= 1(-4 + 6) + 2(-2 + 30) + 3(2 – 20)
= 2 + 56 – 54 = 4 ≠ 0
∴p(A) = 3

Question 2.
Find the rank of the following matrices by row reduction method:

Solution:

The last equivalent matrix is in row echelon form. It has two non-zero rows.
∴ p(A) = 2

The last equivalent matrix is in row echelon form. It has three non-zero rows.
∴ p(A) = 3

The last equivalent matrix is in row echelon form. It has three non-zero rows.
∴ p(A) = 3

Question 3.
Find the inverse of each of the following by Gauss-Jordan method:

Solution:

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:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.
Find the adjoint of the following:

Solution:

Question 2.
Find the inverse (if it exists) of the following.

Solution:
\(\begin{bmatrix} -2 & 4 \\ 1 & -3 \end{bmatrix}\)
|A| = 6 – 4 = 2 ≠ 0
∴ A^-1 exists. A is non singular.


|A| = 2(8-7)-3(6-3)+1(21-12)
= 2 – 9 + 9 = 2 ≠ 0. A^-1 exists.

Question 3.

Solution:
\(\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right]\)
|F(α)| = cos α(cos α – 0) – 0 + sin α(0 + sin α)
= cos²α + sin²α = 1
|f(α)| = 1 ≠ 0. [F(α)]^-1 exists.


[∵ cos (-θ) = cos θ ; sin(-θ) = -sin θ]
from (1) and (2) we have
[F(α)]^-1 = F(-α)

Question 4.
If A = \(\begin{bmatrix} 5 & 3 \\ -1 & -2 \end{bmatrix}\), show that A² – 3A – 7I₂ = O₂. hence find A^-1
Solution:

∴ A² -3A – 7I₂ = O₂
Post multiply this equation by A^-1
A²A^-1 – 3A A^-1 – 7I₂ A^-1 = 0
A – 3I – 7A^-1 = 0
A – 3I = 7 A^-1
A^-1 = \(\frac {1}{7}\) (A – 3I)

Question 5.

Solution:

Question 6.
If A = \(\begin{bmatrix} 8 & -4 \\ -5 & 3 \end{bmatrix}\) verify that A(adj A) = (adj A) A = \(\left| A \right|\)I₂.
Solution:

(1), (2) and (3) ⇒ A (adj A) = (adj A)A = |A| I₂.

Question 7.

Solution:

Question 8.

Solution:

|adj (A)| = 2 (24 – 0) + 4 (- 6 – 14) + 2(0 + 24)
= 48 – 80 + 48 = 16

Question 9.

Solution:

|adj A| = 0 + 2(36 – 18) + 0 = 2(18) = 36

Question 10.

Solution:

Question 11.

Solution:


Hence proved

Question 12.

Solution:

Question 13.

Solution:
Given A × B × C
⇒ A^-1 A × B B^-1 = A^-1 C B^-1
I × I = A^-1 C B^-1
⇒ X = A^-1 CB^-1
let us find A^-1 and B^-1

Question 14.

Solution:


Hence proved.

Question 15.
Decrypt the received encoded message [2 -3] [20 4] with the encryption matrix \(\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1 – 26 to the letters A – Z respectively, and the number 0 to a blank space.
Solution:

So the sequence of decoded row matrics is [8 5] [12 16]
The receiver reads the message as “HELP”.

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