# RD SHARMA Solutions for Class 10 Maths Chapter 10 - Trigonometric Ratios

Page / Exercise

## Chapter 10 - Trigonometric Ratios Exercise Ex. 10.1

Solution 1 (i)

Solution 1 (ii)

Solution 1 (iii)

Solution 1 (iv)

Solution 1 (v)

Solution 1 (vi)

Solution 1 (vii)

Solution 1 (viii)

Solution 1 (ix)

Solution 1 (x)

Solution 1 (xi)

Solution 1 (xii)

Solution 2

In ABC by applying Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7)2
= 576 + 49
= 625
AC =  = 25 cm

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 25

Given:

## Chapter 10 - Trigonometric Ratios Exercise Ex. 10.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26 (i)

Solution 26 (ii)

Solution 26 (iii)

Solution 26 (iv)

Solution 27 (i)

Solution 27 (ii)

Solution 27 (iii)

Solution 28 (i)

Solution 28 (ii)

Solution 29

Solution 30 (i)

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 30 (ii)

Given: tan(A + B) = 1 and

Therefore,

A + B = 45o … (i)

A - B = 30o … (ii)

Adding the two equations, we get

## Chapter 10 - Trigonometric Ratios Exercise Ex. 10.3

Solution 1

Solution 2 (i)

Solution 2 (ii)

Solution 2 (iii)

Solution 2 (iv)

Solution 2 (v)

Solution 2 (vi)

Solution 2 (vii)

Solution 2 (viii)

Solution 2 (ix)

Solution 2 (x)

Solution 2 (xi)

Solution 3

Solution 4

Solution 5

Solution 6(i)

Solution 6(ii)

Solution 7 (i)

Solution 7 (ii)

Solution 7 (iii)

Solution 7 (iv)

Solution 8 (i)

Solution 8 (ii)

Solution 8 (iii)

Solution 8 (iv)

Solution 8 (v)

Solution 9 (i)

Solution 9 (ii)

Solution 9 (iii)

Solution 9 (iv)

Solution 9 (v)

Solution 9 (vi)

Solution 9 (vii)

Solution 9 (viii)

Solution 9 (ix)

Solution 9 (x)

Solution 10

Solution 11 (ii)

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 6 (iii)

Given: A = 90o

For a triangle ABC, A + B + C = 90o

Solution 9 (xi)

Using the identities

Solution 11 (i)

For a triangle ABC, A + B + C = 90o

Solution 18

Given: tan 2A = cot(A - 18o)

As tan x = cot(90o - x), we have

cot(90o - 2A) = cot(A - 18o)

90o - 2A = A - 18o

3A = 108o

Therefore, A = 36o.

## Chapter 10 - Trigonometric Ratios Exercise 10.56

Solution 1

Solution 2

Solution 3

Solution 4

So, the correct option is (a).

Solution 5

Solution 6

## Chapter 10 - Trigonometric Ratios Exercise 10.57

Solution 7

Solution 8

Solution 9

Solution 10

So, the correct option is (d).

Solution 11

So, the correct option is (c).

Solution 12

Solution 13

Solution 14

Solution 15

So, the correct option is (b).

Solution 16

So, the correct option is (a).

Solution 17

So, the correct option is (b).

Solution 18

So, the correct option is (d).

## Chapter 10 - Trigonometric Ratios Exercise 10.58

Solution 19

Solution 20

So, the correct option is (b).

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

So, the correct option is (d).

Solution 28

So, the correct option is (a).

Solution 29

Solution 30

Solution 31

Solution 32

## Chapter 10 - Trigonometric Ratios Exercise 10.59

Solution 33

So, the correct option is (c).

Solution 34

Solution 35

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