Trigonometry Functions – Class XI -Exercise 3.1

1. Find the radian measures corresponding to the following degree

(i) 25˚

(ii) -47˚30’

(iii) 240˚

(iv) 520˚

Solution:

(i) 25˚

We know that 180˚ = π radian

(ii) -47˚30’

-47˚30’ = -471/2 = -95/2 degree

Therefore -47˚30` = -19/72 π radian

(iii) 240˚

We know that 180˚ = π radians

(iv) 520˚

We know that 180˚ = π radian

1. Find the degree measures corresponding to the following radian measures

(use π = 22/7)

(i) 11/16

(ii) -4

(iii) /3

(iv) /6

Solution:

(i) 11/16

We know that π radian = 180˚

11/16 radian = 180/π x 11/16 degree = 54×11/πx4 degree

= 54x11x7/22×4 degree = 315/8 degree

= 39 3/8 degree

= 39˚ + 22’ + 1/2 minutes

= 39˚22’30”

(ii) -4

We know that π radian = 180˚

-4 radian = 180/π x (-4) degree = 180×7(-4)/22 degree

= -2520/11 degree = -2291/11 degree

=-229˚+ 1×60/11 minutes            [1˚ = 60’]

= -229˚+5’+5/11 minutes

= -229˚5’27”

(iii)/3

We know that π radian = 180˚

/3 radian = 180/π x /3 degree = 300˚

(iv) /6

We know that π radian = 180˚

Therefore 7π/6 radian = 180/π x /6 = 210˚

1. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution:

Number of revolutions made by the wheel in 1 minute = 360

Number of revolutions made by the wheel in 1 second = 360/60 = 6

In one complete revolution, the wheel turns an angle of 2π radian.

Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12 π radian

Thus, in one second, the wheel turns an angle of 12π radian.

4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm.

(π = 22/7)

Solution:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then

Therefore, r = 100 cm, l = 22 cm,

we have

θ = 22/100 radian = 180/π x 22/100 degree = 180x7x22/22×100 degree

= 126/10 degree = 123/5 degree = 12˚36’

Thus, the required angle is 12°36′.

1. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Solution:

Diameter of the circle = 40 cm

Radius (r) of the circle = 40/2 cm = 20 cm

Let AB be a chord (length = 20 cm) of the circle.

In ∆OAB,

OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ∆OAB is an equilateral triangle.

θ = 60° = π/3 radians.

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

θ = 1/r

π/3 = AB/20

thus, AB = 20 π/3 cm

Thus, the length of the minor arc of the chord is 20 π/3 cm

1. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Solution:

Let the radii of the two circles be r1 and r2.

Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2.

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

Therefore, l = r1π/3 and l = (r25 π)/12

r1π/3 = (r25 π)/12

⇒ r1= (r25 π)/4

r1/r2 = 5/4

Thus the ratio of the radii is 5:4

1. Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length

(i) 10 cm

(ii) 15 cm

(iii) 21 cm

Solution:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

It is given that r = 75 cm

(i) Here, l = 10 cm

(ii) Here, l = 15 cm