Class 11 Maths Exercise 3.1: Angles and Radians
Overview
This page provides comprehensive Class 11 Ch 3: Trigonometric Functions - Exercise 3.1 Practice. Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3-1. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Trigonometric Functions: Angles
About Class 11 Maths Trigonometric Functions Exercise 3.1
This NCERT Class 11 Maths exercise focuses on Trigonometric Functions: Angles from the chapter Trigonometric Functions. Revise the relevant definitions, standard results, notation, and worked examples before attempting the questions.
Attempt each question independently and write the reasoning in complete steps. Check symbols, substitutions, diagrams, and final conditions carefully, then use the step-by-step solution to identify exactly what needs revision.
Q1
00:00
Find the radian measure corresponding to the following degree measures:
(i) $25^\circ$
(ii) $-47^\circ 30'$
(i) $25^\circ$
(ii) $-47^\circ 30'$
(i) We know that $180^\circ = \pi$ radian.
$25^\circ = 25 \times \frac{\pi}{180} = \frac{5\pi}{36}$ radian.
(ii) $-47^\circ 30' = -(47 + \frac{30}{60})^\circ = -47\frac{1}{2}^\circ = -\frac{95}{2}^\circ$.
Radian measure = $-\frac{95}{2} \times \frac{\pi}{180} = -\frac{19\pi}{72}$ radian.
(i) $\frac{5\pi}{36}$, (ii) $-\frac{19\pi}{72}$
Q2
00:00
Find the degree measure corresponding to the following radian measures (Use $\pi = \frac{22}{7}$):
(i) $\frac{11}{16}$
(ii) $-4$
(i) $\frac{11}{16}$
(ii) $-4$
(i) $\frac{11}{16}$ rad = $\frac{11}{16} \times \frac{180}{\pi}^\circ = \frac{11}{16} \times \frac{180}{22/7}^\circ = \frac{11 \times 180 \times 7}{16 \times 22}^\circ = \frac{315}{8}^\circ$.
$= 39^\circ + \frac{3}{8}^\circ = 39^\circ + (\frac{3}{8} \times 60)' = 39^\circ + 22.5' = 39^\circ 22' 30''$.
(ii) $-4$ rad = $-4 \times \frac{180}{\pi}^\circ = -4 \times \frac{180}{22/7}^\circ = -\frac{2520}{11}^\circ$.
$= -229^\circ + \frac{1}{11}^\circ = -229^\circ + (\frac{1}{11} \times 60)' = -229^\circ + 5' + \frac{5}{11}'$.
$= -229^\circ 5' + (\frac{5}{11} \times 60)'' \approx -229^\circ 5' 27''$.
(i) $39^\circ 22' 30''$, (ii) $-229^\circ 5' 27''$
Q3
00:00
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Number of revolutions in 1 minute = 360.
Number of revolutions in 1 second = $\frac{360}{60} = 6$.
Angle turned in 1 revolution = $2\pi$ radian.
Angle turned in 6 revolutions = $6 \times 2\pi = 12\pi$ radian.
$12\pi$ radians
Q4
00:00
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 $\pi = \frac{22}{7}$).
We use $\theta = \frac{l}{r}$. Here $l = 22$ cm, $r = 100$ cm.
$\theta = \frac{22}{100}$ radian.
In degrees: $\frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times \frac{180}{22/7} = \frac{22}{100} \times \frac{180 \times 7}{22} = \frac{126}{10}^\circ = 12.6^\circ$.
$12.6^\circ = 12^\circ + 0.6 \times 60' = 12^\circ 36'$.
$12^\circ 36'$
Q5
00:00
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Diameter = 40 cm $\Rightarrow$ Radius $r = 20$ cm.
Chord length = 20 cm.
Since radius = chord = 20 cm, the triangle formed by the radii and the chord is equilateral.
Angle subtended at centre $\theta = 60^\circ = \frac{\pi}{3}$ radian.
Arc length $l = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}$ cm.
$\frac{20\pi}{3}$ cm
Q6
00:00
If in two circles, arcs of the same length subtend angles $60^\circ$ and $75^\circ$ at the centre, find the ratio of their radii.
Let radii be $r_1$ and $r_2$. Let arc length be $l$.
$\theta_1 = 60^\circ = \frac{\pi}{3}$ rad. $\theta_2 = 75^\circ = \frac{5\pi}{12}$ rad.
$l = r_1 \theta_1 = r_2 \theta_2 \Rightarrow r_1 \frac{\pi}{3} = r_2 \frac{5\pi}{12}$.
$\frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5}{12} \times \frac{3}{1} = \frac{5}{4}$.
5:4