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.