# RD SHARMA Solutions for Class 11-science Maths Chapter 5 - Trigonometric Functions

Page / Exercise

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 1(i)

Solution 1(ii)
Solution 1(iii)

Solution 1(iv)
Solution 2
Solution 3

Solution 4
Solution 5

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 1(xiii)
Solution 1(xiv)
Solution 2(i)
Solution 2(ii)

Solution 2(iii)

Solution 2(iv)
Solution 2(v)
Solution 2(vi)

Solution 2(vii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 3(v)
Solution 4

Solution 5
Solution 6(i)
Solution 6(ii)
Solution 6(iii)
Solution 7
Solution 8(i)
Solution 8(ii)
Solution 9(i)
Solution 9(ii)
Solution 9(iii)
Solution 9(iv)
Solution 9(v)

## Chapter 5 - Trigonometric Functions Exercise Ex. 5VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

The maximum value for sin(x) is 1 for all x.

Solution 9

Solution 10

The radius of circular wire is 15 cm.

The circumference of circular wire is.

The radius of loop is 120 cm.

The circumference of loop is.

The circumference of a circle covers all the quadrants of the circle i.e. the angle subtended by it at the centre is.

The circular wire is cut and bent so as to lie along the circumstance of the loop.

So, circumference of circular wire covers

of the circumference of the loop.

Therefore the angle subtended by the circular wire is

.

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

So the least value of is 2.

Solution 16

Solution 17

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