10th mathematics exercise questions with answers

# Trigonometry Exercise 13.1 – Class 10

## Trigonometry Exercise 13.1 – Questions:

1. Find sin θ and cos θ for the following:

(i) (ii) (iii) 2.Find the following:

1. If sin x = 3/5 , cosec x = ______________
2. If cos x = 24/25, sec x = _______________
3. If tan x = 7/24 , cot x = _______________
4. If cosec x = 25/15 , sin x = ____________
5. If sin A = 3/5 and cos A = 4/5, then, tan A = __________
6. If cot A = 8/15 and sin A = 15/17, then, cos A = _________

3.Solve:

1. Given tan A = 3/4, find the value of sin A and cos A.
2. Given cot θ = 20/21­, determine cos θ and cosec θ
3. Given tan A = 7/24­, find the other trigonometric ratios of angle A.
4. If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ
5. If 3 tan θ = 1, find sin θ, cos θ and cot θ
6. If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x
7. If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A
8. If 13 sin A = 5 and A is acute, find the va;ue of 5sin A – 2 cos A/tan A
9. If cos θ = 5/13 and θ is acute, find the value of 5tanθ + 12cot θ/5tanθ – 12cot θ
10. IF 13 cos θ – 5 = 0, find sin θ + cos θ/sin θ – cos θ

## Trigonometry Exercise 13.1 – Solutions:

1. Find sinθ and cosθ for the following:

(i) (ii) (iii) Solution: (i)

sin θ = opposite side/hypotensuse = 24/25

cos θ = adjacent side/hypotensuse = 7/25 (ii)

sin θ = opposite side/hypotensuse = 15/25

cos θ = adjacent side/hypotensuse = 20/25 (iii)

sin θ = opposite side/hypotensuse = 10/25

cos θ = adjacent side/hypotensuse = 24/25

2.Find the following:

1. If sin x = 3/5 , cosec x = ______________
2. If cos x = 24/25, sec x = _______________
3. If tan x = 7/24 , cot x = _______________
4. If cosec x = 25/15 , sin x = ____________
5. If sin A = 3/5 and cos A = 4/5, then, tan A = __________
6. If cot A = 8/15 and sin A = 15/17, then, cos A = _________

Solution:

1. If sin x = 3/5 , cosec x = 5/3
2. If cos x = 24/25, sec x = 25/24
3. If tan x = 7/24 , cot x = 24/7
4. If cosec x = 25/15 , sin x = 15/25
5. If sin A = 3/5 and cos A = 4/5, then, tan A = 3/4
6. If cot A = 8/15 and sin A = 15/17, then, cos A = 8/17

1. Solve:

1.Given tan A = 3/4, find the value of sin A and cos A.

Solution:

We know tan A = sin A/cos A = (opposite side/hypotenuse)/ (adjacent side/hypotenuse)

Sin A = opposite side/hypotenuse In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 32 + 42

= 9 + 16

= 25

AB = 5

sin A =  opposite side/hypotenuse  = 3/5

Cos A = adjacent side/hypotenuse = 4/5

1. Given cot θ = 20/21­, determine cos θ and cosec θ

Solution:

We know that, cot θ = 20/21­

cot θ = cos θ/sin θ = (adjacent side/hypotenuse)/( opposite side/hypotenuse) = adjacent side/opposite side In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 202 + 212

= 400 + 441

= 841

AB = 29

sin A =  opposite side/hypotenuse  = 21/29

Cos A = adjacent side/hypotenuse = 20/29

1. Given tan A = 7/24­, find the other trigonometric ratios of angle A.

Solution:

We know tan A = sin A/cos A = (opposite side/hypotenuse)/ (adjacent side/hypotenuse)

= opposite side/adjacent side = 7/24

Sin A = opposite side/hypotenuse In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

= 72 + 242

= 49 + 576

= 625

AB = 25

sin A =  opposite side/hypotenuse  = 7/25

Cos A = adjacent side/hypotenuse = 24/25

tan A = 7/24

sec A = 25/24

cosec A = 25/7

1. If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ

Solution:

sin θ = opposite side/hypotenuse  =  √3/2 In right angled triangle ABC, ∟C = 90˚, A = θ

By Pythagoras theorem,

AB2 = BC2 + CA2

22 = (√3)2 + CA2

4 = 3 + CA2

4 – 3 = CA2

CA = 1

cos θ = 1/2

tan θ = √3/1 = √3

cot θ + cosec θ = 1/√3 + 2/√3 = 1+2/√3 = 3/√3

1. If 3 tan θ = 1, find sin θ, cos θ and cot θ

Solution:

3 tan θ = 1

tan θ = 1/3 = sin θ/cos θ In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

AB2 = (1)2 + 32

= 1 + 9

= 10

AB = √10

Therefore, sin θ = 1/√10

cos θ = 3/√10

cot θ = 3/1 = 3

1. If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x

Solution:

Given, sec x = 2 = hypotenuse/adjacent side = 2/1 In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

22 = BC2 + 12

4 – 1 = BC2

3 = BC2

BC = √3

sin x = √3/2

tan x = √3/1 = √3

cot x = 1/√3

cot x + cosec x = 1/√3 + 2/√3 = 1+2/√3 = 3/√3

1. If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A

Solution:

4 sin A – 3 cos A = 0

sinA/cos A = 3/4

tan A = 3/4 = opposite side/adjacent side In right angled triangle ABC, we have ∟C = 90˚ and A = θ

AB2 = BC2 + CA2

AB2 = 32 + 42

= 9 + 16

= 25

AB = 5 = hypotenuse

sin A = 3/5

cos A = 4/5

sec A = 5/4

cosec A = 5/3

1. If 13 sin A = 5 and A is acute, find the value of 5sin A – 2 cos A/tan A

Solution:

13 sin A = 5

sin A = 5/13 = opposite side/hypotenuse In right angled triangle ABC, we have ∟C = 90˚ and A = acute angle

AB2 = BC2 + CA2

132 = 52 + CA2

169 – 25 = CA2

144 = CA2

12 = CA = adjacent side

We have to find , 5sin A – 2 cos A/tan A

sin A = 5/13

cos A = 12/13

tan A = 5/12

5sin A – 2 cos A/tan A = (5. 5/13 – 2 . 12/13)/(5/12)

5sin A – 2 cos A/tan A = (25/13 24/13)/(5/12)

5sin A – 2 cos A/tan A  =( 1/13)/(5/12)

5sin A – 2 cos A/tan A = 12/65

1. If cos θ = 5/13 and θ is acute, find the value of 5tanθ + 12cot θ/5tanθ – 12cot θ

Solution:

cos θ = 5/13 = Adjacent side/hypotenuse In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB2 = BC2 + CA2

132 = BC2 + 52

169 – 25 =BC2

144 = BC2

12 = BC = opposite side

We have to find, 5tanθ + 12cot θ/5tanθ – 12cot θ

tan θ = 12/5

cot θ = 5/12

5tanθ + 12cot θ/5tanθ – 12cot θ = (5.12/5 + 12.5/12)/(5.12/5 – 12.5/12)

= (12 + 5)/(12 – 5)

= 17/5

1. If 13 cos θ – 5 = 0, find sin θ + cos θ/sin θ – cos θ

Solution:

13 cos θ – 5 = 0

cos θ = 5/13  = adjacent side/hypotenuse In right angled triangle ABC, we have ∟C = 90˚ and A = θ = acute angle

AB2 = BC2 + CA2

132 = BC2 + 52

169 – 25 =BC2

144 = BC2

12 = BC = opposite side

Therefore, sin θ = 12/13

We have to find,  sin θ + cos θ/sin θ – cos θ

sin θ + cos θ/sin θ – cos θ = (12/13 + 5/13)/(12/135/13)

= (17/13)/(7/13)

= 17/7