Trigonometry Exercise 13.1 – Questions:
- Find sin θ and cos θ for the following:
(i)
(ii)
(iii)
2.Find the following:
- If sin x = 3/5 , cosec x = ______________
- If cos x = 24/25, sec x = _______________
- If tan x = 7/24 , cot x = _______________
- If cosec x = 25/15 , sin x = ____________
- If sin A = 3/5 and cos A = 4/5, then, tan A = __________
- If cot A = 8/15 and sin A = 15/17, then, cos A = _________
3.Solve:
- Given tan A = 3/4, find the value of sin A and cos A.
- Given cot θ = 20/21, determine cos θ and cosec θ
- Given tan A = 7/24, find the other trigonometric ratios of angle A.
- If 2 sin θ = √3, find cos θ, tan θ and cot θ + cosec θ
- If 3 tan θ = 1, find sin θ, cos θ and cot θ
- If sec x = 2, then, find sin x, tan x, cot x and cot x + cosec x
- If 4 sin A – 3 cos A = 0, find sin A, cos A, sec A and cosec A
- If 13 sin A = 5 and A is acute, find the va;ue of 5sin A – 2 cos A/tan A
- If cos θ = 5/13 and θ is acute, find the value of 5tanθ + 12cot θ/5tanθ – 12cot θ
- IF 13 cos θ – 5 = 0, find sin θ + cos θ/sin θ – cos θ
Trigonometry Exercise 13.1 – Solutions:
- 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:
- If sin x = 3/5 , cosec x = ______________
- If cos x = 24/25, sec x = _______________
- If tan x = 7/24 , cot x = _______________
- If cosec x = 25/15 , sin x = ____________
- If sin A = 3/5 and cos A = 4/5, then, tan A = __________
- If cot A = 8/15 and sin A = 15/17, then, cos A = _________
Solution:
- If sin x = 3/5 , cosec x = 5/3
- If cos x = 24/25, sec x = 25/24
- If tan x = 7/24 , cot x = 24/7
- If cosec x = 25/15 , sin x = 15/25
- If sin A = 3/5 and cos A = 4/5, then, tan A = 3/4
- If cot A = 8/15 and sin A = 15/17, then, cos A = 8/17
- 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)
= opposite side/adjacent side
Sin A = opposite side/hypotenuse
Cos A = adjacent 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
- 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
- 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
Cos A = adjacent 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
cot A = 24/7
sec A = 25/24
cosec A = 25/7
- 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
- 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
- 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
- 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
- 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
- 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
- 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)/(12/13 – 5/13)
= (17/13)/(7/13)
= 17/7
