Trigonometric Functions
NCERT Class 11 Maths • Exercise 3.11. Convert Degrees to Radians
$$ \text{Radian Measure} = \frac{\pi}{180} \times \text{Degree Measure} $$
Q1: Find radian measures for (i) 25° (ii) -47°30′ (iii) 240° (iv) 520°
(i) $25^\circ$
$= 25 \times \frac{\pi}{180} = \frac{5\pi}{36}$ radians
$= 25 \times \frac{\pi}{180} = \frac{5\pi}{36}$ radians
(ii) $-47^\circ 30’$
First convert minutes to degrees: $30′ = \frac{30}{60}^\circ = \frac{1}{2}^\circ$.
Angle $= -(47 + \frac{1}{2})^\circ = -\frac{95}{2}^\circ$.
In Radians: $-\frac{95}{2} \times \frac{\pi}{180} = -\frac{19\pi}{72}$ radians
First convert minutes to degrees: $30′ = \frac{30}{60}^\circ = \frac{1}{2}^\circ$.
Angle $= -(47 + \frac{1}{2})^\circ = -\frac{95}{2}^\circ$.
In Radians: $-\frac{95}{2} \times \frac{\pi}{180} = -\frac{19\pi}{72}$ radians
(iii) $240^\circ$
$= 240 \times \frac{\pi}{180} = \frac{4\pi}{3}$ radians
$= 240 \times \frac{\pi}{180} = \frac{4\pi}{3}$ radians
(iv) $520^\circ$
$= 520 \times \frac{\pi}{180} = \frac{26\pi}{9}$ radians
$= 520 \times \frac{\pi}{180} = \frac{26\pi}{9}$ radians
2. Convert Radians to Degrees
$$ \text{Degree Measure} = \frac{180}{\pi} \times \text{Radian Measure} $$
(Use $\pi = \frac{22}{7}$)
Q2: Find degree measures for (i) 11/16 (ii) -4 (iii) 5π/3 (iv) 7π/6
(i) $\frac{11}{16}$ rad
$= \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22}$
$= \frac{315}{8}^\circ = 39 \frac{3}{8}^\circ$
Convert fraction: $\frac{3}{8}^\circ = \frac{3}{8} \times 60′ = 22.5′ = 22′ 30”$
Ans: $39^\circ 22′ 30”$
$= \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22}$
$= \frac{315}{8}^\circ = 39 \frac{3}{8}^\circ$
Convert fraction: $\frac{3}{8}^\circ = \frac{3}{8} \times 60′ = 22.5′ = 22′ 30”$
Ans: $39^\circ 22′ 30”$
(ii) $-4$ rad
$= -4 \times \frac{180 \times 7}{22} = – \frac{2520}{11}^\circ$
$= -229 \frac{1}{11}^\circ$
Convert minutes: $\frac{1}{11} \times 60′ = 5 \frac{5}{11}’$
Convert seconds: $\frac{5}{11} \times 60” \approx 27”$
Ans: $-229^\circ 5′ 27”$ (approx)
$= -4 \times \frac{180 \times 7}{22} = – \frac{2520}{11}^\circ$
$= -229 \frac{1}{11}^\circ$
Convert minutes: $\frac{1}{11} \times 60′ = 5 \frac{5}{11}’$
Convert seconds: $\frac{5}{11} \times 60” \approx 27”$
Ans: $-229^\circ 5′ 27”$ (approx)
(iii) $\frac{5\pi}{3}$ rad
$= \frac{5\pi}{3} \times \frac{180}{\pi} = 5 \times 60 = 300^\circ$
$= \frac{5\pi}{3} \times \frac{180}{\pi} = 5 \times 60 = 300^\circ$
(iv) $\frac{7\pi}{6}$ rad
$= \frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^\circ$
$= \frac{7\pi}{6} \times \frac{180}{\pi} = 7 \times 30 = 210^\circ$
3, 4, 5. Applications of $l = r\theta$
Q3: Wheel revolutions problem
Given: 360 revolutions in 1 minute.
Revolutions in 1 second = $\frac{360}{60} = 6$ revolutions.
Angle turned in 1 revolution = $2\pi$ radians.
Angle turned in 6 revolutions = $6 \times 2\pi = 12\pi$ radians.
Ans: $12\pi$ radians
Angle turned in 1 revolution = $2\pi$ radians.
Angle turned in 6 revolutions = $6 \times 2\pi = 12\pi$ radians.
Ans: $12\pi$ radians
Q4: Find degree measure subtended by arc length 22cm (r=100cm)
$$ \theta = \frac{l}{r} \text{ (in radians)} $$
$l = 22, r = 100$.
$\theta = \frac{22}{100}$ radian.
Convert to degrees: $\frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times \frac{180 \times 7}{22}$
$= \frac{126}{10} = 12.6^\circ = 12^\circ 36’$
Ans: $12^\circ 36’$
$\theta = \frac{22}{100}$ radian.
Convert to degrees: $\frac{22}{100} \times \frac{180}{\pi} = \frac{22}{100} \times \frac{180 \times 7}{22}$
$= \frac{126}{10} = 12.6^\circ = 12^\circ 36’$
Ans: $12^\circ 36’$
Q5: Find length of minor arc (Diameter 40cm, Chord 20cm)
Radius $r = \frac{40}{2} = 20$ cm.
Chord length = 20 cm.
Since Radius = Chord = 20 cm, the triangle formed at the center is Equilateral.
Therefore, $\theta = 60^\circ = \frac{\pi}{3}$ radians.
Chord length = 20 cm.
Since Radius = Chord = 20 cm, the triangle formed at the center is Equilateral.
Therefore, $\theta = 60^\circ = \frac{\pi}{3}$ radians.
Arc length $l = r\theta$
$l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}$ cm.
Ans: $\frac{20\pi}{3}$ cm
$l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3}$ cm.
Ans: $\frac{20\pi}{3}$ cm
6, 7. Ratios and Pendulums
Q6: Two circles, same arc length, angles 60° and 75°. Find ratio of radii.
Let arc length be $l$.
$\theta_1 = 60^\circ = \frac{\pi}{3}$ rad.
$\theta_2 = 75^\circ = \frac{5\pi}{12}$ rad.
$\theta_1 = 60^\circ = \frac{\pi}{3}$ rad.
$\theta_2 = 75^\circ = \frac{5\pi}{12}$ rad.
Using $l = r\theta \Rightarrow r = \frac{l}{\theta}$
$\frac{r_1}{r_2} = \frac{l/\theta_1}{l/\theta_2} = \frac{\theta_2}{\theta_1}$
$= \frac{75^\circ}{60^\circ} = \frac{5}{4}$
Ans: 5:4
$\frac{r_1}{r_2} = \frac{l/\theta_1}{l/\theta_2} = \frac{\theta_2}{\theta_1}$
$= \frac{75^\circ}{60^\circ} = \frac{5}{4}$
Ans: 5:4
Q7: Pendulum swing (Length 75cm). Find angle in radians.
Radius $r = 75$ cm. Formula $\theta = \frac{l}{r}$.
(i) Arc 10 cm
$\theta = \frac{10}{75} = \frac{2}{15}$ radian.
$\theta = \frac{10}{75} = \frac{2}{15}$ radian.
(ii) Arc 15 cm
$\theta = \frac{15}{75} = \frac{1}{5}$ radian.
$\theta = \frac{15}{75} = \frac{1}{5}$ radian.
(iii) Arc 21 cm
$\theta = \frac{21}{75} = \frac{7}{25}$ radian.
$\theta = \frac{21}{75} = \frac{7}{25}$ radian.