Practicing the AP Board Solutions Class 11 Maths and Chapter 3 Inter 1st Year Maths Trigonometric Functions Solutions Exercise 3a Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Trigonometric Functions Solutions Exercise 3a
Trigonometric Functions Exercise 3a Solutions
Trigonometric Functions Class 11 Exercise 3a Solutions – Trigonometric Functions 3a Exercise Solutions
Question 1.
Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) -47°30′
(iii) 240°
(iv) 520°
Solution:
(i) Here 180° = π radian
It can be written as
25° = \(\frac{\pi}{180}\) × 25 radian
So, we get \(\frac{5 \pi}{36}\) radian
(ii) Here 1° = 60′
It can be written as -47°30′ = -47\(\frac {1}{2}\) degree
So, we get \(\frac {-95}{2}\) degree
Here 180° = π radian
(iii) Here 180° = π radian
It can be written as
240° = \(\frac{\pi}{180}\) × 240 radian
So, we get \(\frac {4}{3}\)π radian.
(iv) Here 180° = π radian
It can be written as
520° = \(\frac{\pi}{180}\) × 520 radian
So, we get \(\frac {26}{9}\)π radian
Question 2.
Find the degree measure corresponding to the following radian measures.
(Use π = \(\frac {22}{7}\))
(i) \(\frac {11}{16}\)
(ii) -4
(iii) \(\frac{5 \pi}{3}\)
(iv) \(\frac{7 \pi}{6}\)
Solution:
Question 3.
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution:
It is given that
No. of revolutions made by the wheel in 1 minute = 360
1 second = \(\frac {360}{60}\) = 6
We know that
The wheel turns an angle of 2π radians in one complete revolution.
In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12π radian.
Therefore, in one second, the wheel turns an angle of 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:
Consider a circle of radius r units with 1 unit as the arc length, which subtends an angle θ radians at the centre.
θ = \(\frac {1}{r}\)
Here r = 100 cm, l = 22 cm
Therefore, the required angle is 12°36′.
Question 5.
In a circle of a diameter of 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Solution:
The dimensions of the circle are
Diameter = 40 cm
Radius = 20 cm
Consider AB to be the chord of the circle
i.e., length = 20 cm
In ∆OAB,
Radius of circle = OA = OB = 20 cm
Similarly AB = 20 cm
Hence, ∆OAB is an equilateral triangle.
θ = 60° = \(\frac{\pi}{3}\) radian a circle of radius r unit,
If an arc of length a unit subtends an angle θ radian at the centre, we get θ = \(\frac {1}{r}\).
\(\frac{\pi}{3}=\frac{\overparen{\mathrm{AB}}}{20} \Rightarrow \overparen{\mathrm{AB}}=20 \frac{\pi}{3} \mathrm{~cm}\)
Therefore, the length of the minor arc of the chord is 20\(\frac{\pi}{3}\) cm.
Question 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:
Consider r1 and r2 as the radii of the two circles.
Let an arc of length 1 subtend an angle of 60° at the centre of the circle of radius r1,
and an arc of length 1 subtend an angle of 75° at the centre of the circle of radius r2.
Here 60° = \(\frac{\pi}{3}\) radian and 75° = \(\frac{5\pi}{12}\) radian.
In a circle of radius r units, if an arc of length l units subtends an angle θ radians at the centre, we get,
Therefore, the ratio of the radii is 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.
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Solution:
In a circle of radius r units, if an arc of length 1 unit subtends an angle θ radians at the centre, then θ = \(\frac {l}{r}\)
We know that r = 75 cm
(i) l = 10 cm
So, we get θ = \(\frac {10}{75}\) radian
By further simplification
θ = \(\frac {2}{15}\) radian.
(ii) l = 15 cm
So, we get θ = \(\frac {15}{75}\) radian
By further simplification
θ = \(\frac {1}{5}\) radian.
(iii) l = 21 cm
So, we get θ = \(\frac {21}{75}\) radian
By further simplification
θ = \(\frac {7}{25}\) radian.