# Class 11 Maths NCERT Solutions for Chapter 3 Trigonometric Functions Exercise 3.1

### Trigonometric Functions Exercise 3.1 Solutions

**1. Find the radian measures corresponding to the following degree measures:(i) 25° (ii) –47° 30' (iii) 240° (iv) 520°**

**Solution**

(i) 25°

We know that 180° = Ï€ radian

∴ 25° = Ï€/180° ×25 radian = 5Ï€/36 radian

(ii) -47° 30'

- 47° 30' = -47.5 degree **[1° = 60']**

= -95/2 degree

Since, 180° = Ï€ radian

(iii) 240°

We know that 180° = Ï€ radian

∴ 240° = (Ï€/180) ×240 radian = (4/3)Ï€ radian

(iv) 520°

We know that 180° = Ï€ radian

∴ 520° = Ï€/180 ×520 radian = 26Ï€/9 radian

**2. Find the degree measures corresponding to the following radian measures (use Ï€ = 22/7). (i) 11/16 (ii) -4(iii) 5Ï€/3 (iv) 7Ï€/6 **

**Solution**

(i) 11/16

We know that Ï€ radian = 180°

(ii) -4

We know that Ï€ radian = 180°

(iii) 5Ï€/3

We know that Ï€ radian = 180°

∴ 5Ï€/3 radian = 180/Ï€ × 5Ï€/3 degree = 300°

(iv) 7Ï€/6

We know that Ï€ radian = 180°

∴ 7Ï€/6 radian = 180/Ï€ × 7Ï€/6 = 210°

**3. 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 2Ï€ radian.

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

12 Ï€ radian

Thus, in one second, the wheel turns an angle of 12Ï€ 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 (use Ï€ = 22/7) . **

**Solution**

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

Î¸ = 1/r

Therefore, for = 100 cm, l = 22 cm, we have

Thus, the required angle is 12° 36'.

**5. 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 Î”OAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, Î”OAB is an equilateral triangle.

∴Î¸ = 60° = Ï€/3 radian

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

Thus, the length of the minor arc of the chord is 20Ï€/3 cm.

**6. 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 and . Let an arc of length *l* subtend an angle of 60° at the centre of the circle of radius *r*_{1}, while let an arc of length *l *subtend an angle of 75° at the centre of the circle of radius *r*_{2}.

Now, 60° = Ï€/3 radian and 75° = 5Ï€/12 radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle Î¸ radian at the centre, then Î¸ = 1/r or *l* = rÎ¸ .

Thus, the ratio of the radii is 5 : 4.

**7. 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 Î¸ radian at the centre, then Î¸ = l/r.

It is given that r = 75 cm

(i) Here, l = 10 cm

Î¸ = 10/75 radian = 2/15 radian

(ii) Here, l = 15 cm

Î¸ = 15/75 radian = 1/5 radian

(iii) Here, l = 21 cm

Î¸ = 21/75 radian = 7/25 radian