Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1

Question 1.

Find the radian measures corresponding to the following degree measures?

- 25°
- – 47°30′
- 240°
- 520°

Solution:

1. We have:

180° = π radians

∴ 25° = \(\frac{π}{180}\) × 25 radians = \(\frac{5π}{36}\) radians.

2. 60′ = 1°

∴ 30′ = \(\frac{30°}{60}\) = \(\frac{1°}{2}\)

∴ 47°30′ = (47 + \(\frac{1}{2}\)) degrees = \(\frac{95}{2}\) degrees

Now, 180° = π radians

So, – \(\frac{95°}{2}\) = \(\frac{- π}{180}\) × \(\frac{95}{2}\) radians = \(\frac{- 19π}{72}\) radians.

3. 180° = π radians.

∴ 240° = \(\frac{π}{180}\) × 240 radians

= \(\frac{4π}{3}\) radians.

4. 180° = π radians

∴ 520° = \(\frac{π}{180}\) × 520 radians = \(\frac{26π}{9}\) radians.

Question 2.

Find the degree measures corresponding to the following radian measures. [Use π = \(\frac{22}{7}\)]

(i) \(\frac{11}{16}\)

(ii) – 4

(iii) \(\frac{5π}{3}\)

(iv) \(\frac{7π}{6}\)

Solution:

(i) π radians = \(\frac{22}{7}\) radians = 180°

∴ \(\frac{11}{16}\) radians = \(\frac{180}{22}\) × 7 × \(\frac{11}{16}\) degree

= \(\frac{315}{8}\) degrees = 39 \(\frac{3}{8}\) degrees

= 39°22’30”.

Note: \(\frac{3°}{8}\) = \(\frac{3}{8}\) × 60′ = \(\frac{45}{2}\) = 22’30”.

(ii)

(iii) \(\frac{5π}{3}\) = \(\frac{5}{3}\) × 180° = 300°.

(iv) \(\frac{7π}{6}\) = \(\frac{7}{6}\) × 180° = 210°.

Question 3.

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

Solution:

Angle rotated in one revolution = 2π radians

∴ Angle rotated in 360 revolutions = 360 × 2π radians

⇒ Angle turned in one minute or 60 sec = 360 × 2π.

Hence, angle turned in 1 sec = \(\frac{360 × 2π}{60}\).

= 12π radians.

Question 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 π = \(\frac{22}{7}\))

Solution:

We know that:

l = rθ,

where l = length of arc = 20 cm,

r = radius of circle = 100 cm

and θ = angle subtended at the centre

Question 5.

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

Solution:

Since radius = length of chord = 20 cm, so

∆ OAB is equilateral triangle

⇒ θ = 60°.

Now, l = rθ

So, l = 20 × 60° × \(\frac{π}{180°}\) = \(\frac{20π}{3}\)

Thus, l = \(\frac{20π}{3}\) cm.

Question 6.

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

Solution:

Since l is same for both the circles, therefore

\(\frac{π}{3}\)r_{1} = \(\frac{5π}{12}\)r_{2}.

⇒ r_{1} : r_{2} = 5 : 4.

Question 7.

Find the angle in radians through which a pendulum swings, if its length is 75 cm and the tip describes an arc of length:

- 10 cm
- 15 cm
- 21 cm

Solution:

1. r = 75 cm,

l = 10 cm,

θ = ?

⇒ θ = \(\frac{l}{r}\) = \(\frac{10}{75}\) radians.

= \(\frac{2}{15}\) radians

2. r = 75 cm,

l = 15 cm,

∴ θ = \(\frac{l}{r}\) = \(\frac{15}{75}\) radians = \(\frac{1}{5}\) radians.

3. r = 75 cm,

l = 21 cm,

∴ θ = \(\frac{l}{r}\) = \(\frac{21}{75}\) radians = \(\frac{7}{25}\) radians.