NCERT Solutions: Exercise 3.1

Quick Concept Recap

Before jumping into the solutions, here are the key formulas you need for this exercise:

1. Relation between Degree and Radian:

• $\pi \text{ radian} = 180^{\circ}$

• $1 \text{ radian} = \frac{180^{\circ}}{\pi}$

• $1^{\circ} = \frac{\pi}{180} \text{ radian}$

2. Arc Length Formula:

   $$ l = r \theta $$

   Where $l$ is the length of the arc, $r$ is the radius, and $\theta$ is the angle in radians.

Detailed Solutions

Question 1: Find the radian measures corresponding to the following degree measures:

(i) $25^{\circ}$ (ii) $-47^{\circ}30'$ (iii) $240^{\circ}$ (iv) $520^{\circ}$

Solution:

We know that $180^{\circ} = \pi \text{ radian}$. Therefore, $1^{\circ} = \frac{\pi}{180} \text{ radian}$.

(i) $25^{\circ}$

$$ 25^{\circ} = 25 \times \frac{\pi}{180} \text{ radian} = \frac{5\pi}{36} \text{ radian} $$

(ii) $-47^{\circ}30'$

Step 1: Convert minutes to degrees.

$$ 30' = \frac{30}{60}^{\circ} = \frac{1}{2}^{\circ} $$

$$ \implies -47^{\circ}30' = -\left(47 \frac{1}{2}\right)^{\circ} = -\frac{95}{2}^{\circ} $$

Step 2: Convert to radians.

$$ -\frac{95}{2}^{\circ} = -\frac{95}{2} \times \frac{\pi}{180} \text{ radian} = -\frac{19\pi}{72} \text{ radian} $$

(iii) $240^{\circ}$

$$ 240^{\circ} = 240 \times \frac{\pi}{180} \text{ radian} = \frac{4\pi}{3} \text{ radian} $$

(iv) $520^{\circ}$

$$ 520^{\circ} = 520 \times \frac{\pi}{180} \text{ radian} = \frac{26\pi}{9} \text{ radian} $$

Question 2: Find the degree measures corresponding to the following radian measures (Use $\pi = \frac{22}{7}$):

(i) $\frac{11}{16}$ (ii) $-4$ (iii) $\frac{5\pi}{3}$ (iv) $\frac{7\pi}{6}$

Solution:

We know that $\pi \text{ radian} = 180^{\circ}$. Therefore, $1 \text{ radian} = \frac{180^{\circ}}{\pi}$.

(i) $\frac{11}{16}$

$$ \frac{11}{16} \text{ radian} = \frac{11}{16} \times \frac{180}{\pi}^{\circ} $$

$$ = \frac{11}{16} \times \frac{180}{22/7}^{\circ} = \frac{11 \times 180 \times 7}{16 \times 22}^{\circ} = \frac{315}{8}^{\circ} $$

Convert fraction to minutes and seconds:

$$ \frac{315}{8}^{\circ} = 39 \frac{3}{8}^{\circ} = 39^{\circ} + \left(\frac{3}{8} \times 60\right)' $$

$$ = 39^{\circ} + 22.5' = 39^{\circ} 22' + \left(0.5 \times 60\right)'' = 39^{\circ} 22' 30'' $$

Answer: $39^{\circ} 22' 30''$

(ii) $-4$

$$ -4 \text{ radian} = -4 \times \frac{180}{\pi}^{\circ} = -4 \times \frac{180 \times 7}{22}^{\circ} $$

$$ = -\frac{2520}{11}^{\circ} $$

Convert fraction:

$$ -\frac{2520}{11}^{\circ} = -229 \frac{1}{11}^{\circ} = -229^{\circ} + \left(\frac{1}{11} \times 60\right)' $$

$$ = -229^{\circ} + 5 \frac{5}{11}' = -229^{\circ} 5' + \left(\frac{5}{11} \times 60\right)'' \approx -229^{\circ} 5' 27'' $$

Answer: $\approx -229^{\circ} 5' 27''$

(iii) $\frac{5\pi}{3}$

$$ \frac{5\pi}{3} \text{ radian} = \frac{5\pi}{3} \times \frac{180}{\pi}^{\circ} = 5 \times 60^{\circ} = 300^{\circ} $$

(iv) $\frac{7\pi}{6}$

$$ \frac{7\pi}{6} \text{ radian} = \frac{7\pi}{6} \times \frac{180}{\pi}^{\circ} = 7 \times 30^{\circ} = 210^{\circ} $$

Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution:

• Revolutions per minute = 360

• Revolutions per second = $360 / 60 = 6$

• Angle in 1 revolution = $2\pi$ radians

• Angle in 6 revolutions = $6 \times 2\pi = 12\pi$ radians

Answer: $12\pi$ 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 $\pi = \frac{22}{7}$).

Solution:

Using the formula $\theta = \frac{l}{r}$:

$$ \theta = \frac{22}{100} \text{ radian} = \frac{11}{50} \text{ radian} $$

Convert to degrees:

$$ \frac{11}{50} \times \frac{180}{\pi}^{\circ} = \frac{11}{50} \times \frac{180}{22/7}^{\circ} $$

$$ = \frac{11 \times 180 \times 7}{50 \times 22}^{\circ} = \frac{126}{10}^{\circ} = 12.6^{\circ} $$

Convert decimal:

$$ 12.6^{\circ} = 12^{\circ} + (0.6 \times 60)' = 12^{\circ} 36' $$

Answer: $12^{\circ} 36'$

Question 5: 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.

Solution:

• Radius $r = \frac{40}{2} = 20$ cm. Chord = 20 cm.

• Since Radius = Chord = 20 cm, the triangle formed is equilateral.

• Angle $\theta = 60^{\circ} = \frac{\pi}{3}$ radians.

• Arc length $l = r\theta$:

  $$ l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \text{ cm} $$

Answer: $\frac{20\pi}{3}$ cm

Question 6: 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.

Solution:

Given: $\theta_1 = 60^{\circ} = \frac{\pi}{3}$, $\theta_2 = 75^{\circ} = \frac{5\pi}{12}$, and $l_1 = l_2 = l$.

Using $r = \frac{l}{\theta}$:

$$ \frac{r_1}{r_2} = \frac{l/\theta_1}{l/\theta_2} = \frac{\theta_2}{\theta_1} $$

$$ = \frac{5\pi/12}{\pi/3} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{5}{4} $$

Answer: 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:

Here, radius $r = 75$ cm. Use $\theta = \frac{l}{r}$.

(i) $l = 10$ cm

$$ \theta = \frac{10}{75} = \frac{2}{15} \text{ radian} $$

(ii) $l = 15$ cm

$$ \theta = \frac{15}{75} = \frac{1}{5} \text{ radian} $$

(iii) $l = 21$ cm

$$ \theta = \frac{21}{75} = \frac{7}{25} \text{ radian} $$

Expert Tips & Common Mistakes

• Radians is the Default: Always convert degrees to radians before using formulas like $l = r\theta$.

• Conversion Trick:

  • Degrees $\to$ Radians: Multiply by $\frac{\pi}{180}$.

  • Radians $\to$ Degrees: Multiply by $\frac{180}{\pi}$.

• Units Matter: Always check if the question asks for the answer in degrees or radians. Losing marks for missing units is common!