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?

- 25Â°
- – 47Â°30′
- 240Â°
- 520Â°

Solution:

1. We have:

180Â° = Ï€ radians

âˆ´ 25Â° = \(\frac{Ï€}{180}\) Ã— 25 radians = \(\frac{5Ï€}{36}\) radians.

2. 60′ = 1Â°

âˆ´ 30′ = \(\frac{30Â°}{60}\) = \(\frac{1Â°}{2}\)

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

Now, 180Â° = Ï€ radians

So, – \(\frac{95Â°}{2}\) = \(\frac{- Ï€}{180}\) Ã— \(\frac{95}{2}\) radians = \(\frac{- 19Ï€}{72}\) radians.

3. 180Â° = Ï€ radians.

âˆ´ 240Â° = \(\frac{Ï€}{180}\) Ã— 240 radians

= \(\frac{4Ï€}{3}\) radians.

4. 180Â° = Ï€ radians

âˆ´ 520Â° = \(\frac{Ï€}{180}\) Ã— 520 radians = \(\frac{26Ï€}{9}\) radians.

Question 2.

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

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

(ii) – 4

(iii) \(\frac{5Ï€}{3}\)

(iv) \(\frac{7Ï€}{6}\)

Solution:

(i) Ï€ radians = \(\frac{22}{7}\) radians = 180Â°

âˆ´ \(\frac{11}{16}\) radians = \(\frac{180}{22}\) Ã— 7 Ã— \(\frac{11}{16}\) degree

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

= 39Â°22’30”.

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

(ii)

(iii) \(\frac{5Ï€}{3}\) = \(\frac{5}{3}\) Ã— 180Â° = 300Â°.

(iv) \(\frac{7Ï€}{6}\) = \(\frac{7}{6}\) Ã— 180Â° = 210Â°.

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 = 2Ï€ radians

âˆ´ Angle rotated in 360 revolutions = 360 Ã— 2Ï€ radians

â‡’ Angle turned in one minute or 60 sec = 360 Ã— 2Ï€.

Hence, angle turned in 1 sec = \(\frac{360 Ã— 2Ï€}{60}\).

= 12Ï€ 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 Ï€ = \(\frac{22}{7}\))

Solution:

We know that:

l = rÎ¸,

where l = length of arc = 20 cm,

r = radius of circle = 100 cm

and Î¸ = 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

âˆ† OAB is equilateral triangle

â‡’ Î¸ = 60Â°.

Now, l = rÎ¸

So, l = 20 Ã— 60Â° Ã— \(\frac{Ï€}{180Â°}\) = \(\frac{20Ï€}{3}\)

Thus, l = \(\frac{20Ï€}{3}\) cm.

Question 6.

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

Solution:

Since l is same for both the circles, therefore

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

â‡’ 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,

Î¸ = ?

â‡’ Î¸ = \(\frac{l}{r}\) = \(\frac{10}{75}\) radians.

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

2. r = 75 cm,

l = 15 cm,

âˆ´ Î¸ = \(\frac{l}{r}\) = \(\frac{15}{75}\) radians = \(\frac{1}{5}\) radians.

3. r = 75 cm,

l = 21 cm,

âˆ´ Î¸ = \(\frac{l}{r}\) = \(\frac{21}{75}\) radians = \(\frac{7}{25}\) radians.