Class X Chapter 5 – Trigonometric Ratios Maths
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Exercise 5.1
1. In each of the following one of the six trigonometric ratios is given. Find the values of the
other trigonometric ratios.
Sol:
2
(i) Sin A = 3
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
We know that Sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Let us Consider a right angled ∆le ABC.
By applying Pythagorean theorem we get
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
9 = 𝑥2 + 4
𝑥2 = 9 - 4
x = √5
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
We know that cos = and
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
tan𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
√5
So, cos𝜃 = ;
3
1 3
sec = =
𝑐𝑜𝑠𝜃 √5
2
tan𝜃 = ;
√5
1 √5
cot = 𝑡𝑎𝑛𝜃 = 2
1 3
cosec𝜃 = 𝑠𝑖𝑛𝜃 = 2
(ii)
4
Cos A = 5
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
We know that cos𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Let us consider a right angled ∆le ABC.
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,Class X Chapter 5 – Trigonometric Ratios Maths
______________________________________________________________________________
Let opposite side BC = x.
By applying pythagorn’s theorem, we get
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
25 = x + 16
x = 25 - 16 = 9
x = √9 = 3
4
We know that cosA = 5
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 3
sinA = =5
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 3
tanA = =4
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
1
1 3 5
cosecA = 𝑠𝑖𝑛𝐴 = =3
5
1
1 4 5
secA = = =4
𝑐𝑜𝑠𝐴 5
1
1 3 4
cotA = 𝑡𝑎𝑛𝐴 = =3
4
(iii)
tan𝜃 = 11.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know that tan𝜃 = =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
Consider a right angled ∆ ABC. le
Let hypotenuse AC = x, by applying Pythagoras theorem
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
𝑥 2 = 112 + 12
𝑥 2 = 121 + 1
𝑥 = √122
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know that sin𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 √122
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
cos𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 √122
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,Class X Chapter 5 – Trigonometric Ratios Maths
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1
1 11 √122
cosec𝜃 = 𝑠𝑖𝑛𝜃 = =
√122 11
1 1/1
sec𝜃 = = = √122
𝑐𝑜𝑠𝜃 √122
1 1 1
cot𝜃 = 𝑡𝑎𝑛𝜃 = 11 = 11
(iv)
11
Sin 𝜃 = 5
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know Sin 𝜃 = = 15
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Consider right angled ∆le ACB.
A
II I�
c e 6B
'lC..
Let x = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
By applying Pythagoras
𝐴𝐵 2 = 𝐴𝐶 2 + 𝐵𝐶 2
225 = 121+𝑥 2
𝑥 2 = 225 -121
𝑥 2 = 104
𝑥 = √104
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 104
cos = = √ 15
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
tan = =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 √104
1 15
cosec𝜃 = 𝑠𝑖𝑛𝜃 = 11
1 15
sec = =
𝑐𝑜𝑠𝜃 √104
1 √104
cot = 𝑡𝑎𝑛𝜃 = 11
(v)
5
tanα = 12
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 5
We know that tanα= = 12
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
Now consider a right angled ∆le ABC.
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, Class X Chapter 5 – Trigonometric Ratios Maths
______________________________________________________________________________
A
c
Let x = hypotenuse .By applying Pythagoras theorem
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
𝑥 2 = 52 + 122
𝑥 2 = 25 + 144 = 169
𝑥 = 13
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 5
sinα = = 13
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 12
cosα = = 13
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
1 12
cotα = 𝑡𝑎𝑛α = 15
1 1/5 13
cosecα = 𝑠𝑖𝑛𝛼 = =
13 5
1
1 12 13
secα = = = 12 .
𝑐𝑜𝑠𝛼 13
(vi)
√3
Sin 𝜃 = 2
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 √3
We know Sin 𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
Now consider right angled ∆le ABC.
8
Let x = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
By applying Pythagoras
𝐴𝐵 2 = 𝐴𝐶 2 + 𝐵𝐶 2
4 = 3+𝑥 2
𝑥2 = 4 − 3
𝑥2 = 1
𝑥 =1
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
cos = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 √3
tan = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
= 1
= √3
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Exercise 5.1
1. In each of the following one of the six trigonometric ratios is given. Find the values of the
other trigonometric ratios.
Sol:
2
(i) Sin A = 3
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
We know that Sin 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Let us Consider a right angled ∆le ABC.
By applying Pythagorean theorem we get
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
9 = 𝑥2 + 4
𝑥2 = 9 - 4
x = √5
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
We know that cos = and
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
tan𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
√5
So, cos𝜃 = ;
3
1 3
sec = =
𝑐𝑜𝑠𝜃 √5
2
tan𝜃 = ;
√5
1 √5
cot = 𝑡𝑎𝑛𝜃 = 2
1 3
cosec𝜃 = 𝑠𝑖𝑛𝜃 = 2
(ii)
4
Cos A = 5
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
We know that cos𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Let us consider a right angled ∆le ABC.
______________________________________________________________________________
,Class X Chapter 5 – Trigonometric Ratios Maths
______________________________________________________________________________
Let opposite side BC = x.
By applying pythagorn’s theorem, we get
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
25 = x + 16
x = 25 - 16 = 9
x = √9 = 3
4
We know that cosA = 5
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 3
sinA = =5
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 3
tanA = =4
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
1
1 3 5
cosecA = 𝑠𝑖𝑛𝐴 = =3
5
1
1 4 5
secA = = =4
𝑐𝑜𝑠𝐴 5
1
1 3 4
cotA = 𝑡𝑎𝑛𝐴 = =3
4
(iii)
tan𝜃 = 11.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know that tan𝜃 = =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
Consider a right angled ∆ ABC. le
Let hypotenuse AC = x, by applying Pythagoras theorem
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
𝑥 2 = 112 + 12
𝑥 2 = 121 + 1
𝑥 = √122
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know that sin𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 √122
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
cos𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 √122
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,Class X Chapter 5 – Trigonometric Ratios Maths
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1
1 11 √122
cosec𝜃 = 𝑠𝑖𝑛𝜃 = =
√122 11
1 1/1
sec𝜃 = = = √122
𝑐𝑜𝑠𝜃 √122
1 1 1
cot𝜃 = 𝑡𝑎𝑛𝜃 = 11 = 11
(iv)
11
Sin 𝜃 = 5
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
We know Sin 𝜃 = = 15
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
Consider right angled ∆le ACB.
A
II I�
c e 6B
'lC..
Let x = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
By applying Pythagoras
𝐴𝐵 2 = 𝐴𝐶 2 + 𝐵𝐶 2
225 = 121+𝑥 2
𝑥 2 = 225 -121
𝑥 2 = 104
𝑥 = √104
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 104
cos = = √ 15
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 11
tan = =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 √104
1 15
cosec𝜃 = 𝑠𝑖𝑛𝜃 = 11
1 15
sec = =
𝑐𝑜𝑠𝜃 √104
1 √104
cot = 𝑡𝑎𝑛𝜃 = 11
(v)
5
tanα = 12
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 5
We know that tanα= = 12
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
Now consider a right angled ∆le ABC.
______________________________________________________________________________
, Class X Chapter 5 – Trigonometric Ratios Maths
______________________________________________________________________________
A
c
Let x = hypotenuse .By applying Pythagoras theorem
𝐴𝐶 2 = 𝐴𝐵 2 + 𝐵𝐶 2
𝑥 2 = 52 + 122
𝑥 2 = 25 + 144 = 169
𝑥 = 13
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 5
sinα = = 13
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 12
cosα = = 13
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
1 12
cotα = 𝑡𝑎𝑛α = 15
1 1/5 13
cosecα = 𝑠𝑖𝑛𝛼 = =
13 5
1
1 12 13
secα = = = 12 .
𝑐𝑜𝑠𝛼 13
(vi)
√3
Sin 𝜃 = 2
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 √3
We know Sin 𝜃 = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
Now consider right angled ∆le ABC.
8
Let x = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
By applying Pythagoras
𝐴𝐵 2 = 𝐴𝐶 2 + 𝐵𝐶 2
4 = 3+𝑥 2
𝑥2 = 4 − 3
𝑥2 = 1
𝑥 =1
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 1
cos = =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 2
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 √3
tan = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
= 1
= √3
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