GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1

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?

1. 25°
2. – 47°30′
3. 240°
4. 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
$\frac{π}{3}$r₁ = $\frac{5π}{12}$r₂.
⇒ r₁ : r₂ = 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:

1. 10 cm
2. 15 cm
3. 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.