1. Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) –47° 30′ (iii) 240° (iv) 520°

Concept:

We know the relation between degree and radian measure is \(180^\circ = \pi\) radians. Therefore, to convert degrees to radians, we use the formula:

\( \text{Radian measure} = \frac{\pi}{180} \times \text{Degree measure} \)

(i) 25°

Solution: We have, $$ 25^\circ = \frac{\pi}{180} \times 25 \text{ radians} = \frac{5\pi}{36} \text{ radians} $$ Thus, the radian measure of 25° is \( \frac{5\pi}{36} \).

(ii) –47° 30′

Solution: First, we convert the angle entirely into degrees. We know, \( 60′ = 1^\circ \), so \( 30′ = \frac{30}{60}^\circ = 0.5^\circ \).
So, \( –47^\circ 30′ = -(47 + 0.5)^\circ = -47.5^\circ = -\frac{95}{2}^\circ \).
Now, convert to radians: $$ -\frac{95}{2}^\circ = \frac{\pi}{180} \times \left(-\frac{95}{2}\right) \text{ radians} = -\frac{19\pi}{72} \text{ radians} $$ Thus, the radian measure of –47° 30′ is \( -\frac{19\pi}{72} \).

(iii) 240°

Solution: We have, $$ 240^\circ = \frac{\pi}{180} \times 240 \text{ radians} = \frac{4\pi}{3} \text{ radians} $$ Thus, the radian measure of 240° is \( \frac{4\pi}{3} \).

(iv) 520°

Solution: We have, $$ 520^\circ = \frac{\pi}{180} \times 520 \text{ radians} = \frac{26\pi}{9} \text{ radians} $$ Thus, the radian measure of 520° is \( \frac{26\pi}{9} \).

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

Concept:

To convert radians to degrees, we use the formula:

\( \text{Degree measure} = \frac{180}{\pi} \times \text{Radian measure} \)

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

Solution: We have, $$ \frac{11}{16} \text{ radians} = \left(\frac{180}{\pi} \times \frac{11}{16}\right)^\circ = \left(\frac{180 \times 7}{22} \times \frac{11}{16}\right)^\circ = \left(\frac{315}{8}\right)^\circ $$ Now convert the fraction: $$ \left(\frac{315}{8}\right)^\circ = 39 \frac{3}{8}^\circ = 39^\circ + \left(\frac{3}{8} \times 60\right)’ = 39^\circ + 22.5′ = 39^\circ + 22′ + (0.5 \times 60)” $$ Thus, the degree measure is 39° 22′ 30″.

(ii) –4

Solution: We have, $$ -4 \text{ radians} = \left(\frac{180}{\pi} \times -4\right)^\circ = \left(\frac{180 \times 7}{22} \times -4\right)^\circ = -\left(\frac{2520}{11}\right)^\circ $$ Now convert the fraction: $$ -\left(\frac{2520}{11}\right)^\circ = -229 \frac{1}{11}^\circ = – \left( 229^\circ + \left(\frac{1}{11} \times 60\right)’ \right) = – \left( 229^\circ + 5\frac{5}{11}’ \right) $$ $$ \approx – (229^\circ 5′ 27”) $$ Thus, the degree measure is approximately –229° 5′ 27″.

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

Solution: We have, $$ \frac{5\pi}{3} \text{ radians} = \left(\frac{180}{\pi} \times \frac{5\pi}{3}\right)^\circ = (60 \times 5)^\circ = 300^\circ $$ Thus, the degree measure is 300°.

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

Solution: We have, $$ \frac{7\pi}{6} \text{ radians} = \left(\frac{180}{\pi} \times \frac{7\pi}{6}\right)^\circ = (30 \times 7)^\circ = 210^\circ $$ Thus, the degree measure is 210°.

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

Given: Number of revolutions in 1 minute (60 seconds) = 360.

To Find: The angle in radians the wheel turns in one second.

Solution:

First, find revolutions per second: $$ \text{Revolutions per second} = \frac{360}{60} = 6 $$
We know that \( 1 \text{ revolution} = 2\pi \text{ radians} \).
Angle turned in one second = \( 6 \times 2\pi = 12\pi \) radians.
The wheel turns through 12π radians in one second.

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 \( \pi = \frac{22}{7} \)).

Given: Radius \( r = 100 \) cm, Arc length \( l = 22 \) cm.

To Find: The angle \( \theta \) in degrees.

Formula: \( \theta = \frac{l}{r} \) (gives angle in radians).

Solution:

Angle in radians: $$ \theta = \frac{22}{100} \text{ radians} $$
Convert to degrees: $$ \left( \frac{180}{\pi} \times \frac{22}{100} \right)^\circ = \left( \frac{180 \times 7}{22} \times \frac{22}{100} \right)^\circ = \left( \frac{126}{10} \right)^\circ = 12.6^\circ $$
Convert fractional part to minutes: $$ 12.6^\circ = 12^\circ + (0.6 \times 60)’ = 12^\circ + 36′ $$
The required angle is 12° 36′.

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.

Given: Diameter = 40 cm, Chord length = 20 cm.

To Find: Length of the minor arc.

Solution:

Radius \( r = \frac{40}{2} = 20 \) cm.
Consider the triangle formed by the chord and two radii to the chord’s endpoints. Its sides are 20 cm, 20 cm, and 20 cm. Thus, it is an equilateral triangle.
The angle at the centre is \( \theta = 60^\circ \).
Convert angle to radians: $$ \theta = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians} $$
Find arc length using \( l = r\theta \): $$ l = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \text{ cm} $$
The length of the minor arc is \( \frac{20\pi}{3} \) cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Given: Let radii be \(r_1, r_2\). Let angles be \( \theta_1 = 60^\circ, \theta_2 = 75^\circ \). Arc lengths are equal, \( l_1 = l_2 = l \).

To Find: The ratio \( r_1 : r_2 \).

Solution:

From \( l = r\theta \), we have \( l = r_1\theta_1 \) and \( l = r_2\theta_2 \).
Therefore, \( r_1\theta_1 = r_2\theta_2 \).
The ratio is: $$ \frac{r_1}{r_2} = \frac{\theta_2}{\theta_1} $$
Using the degree values directly: $$ \frac{r_1}{r_2} = \frac{75}{60} = \frac{5}{4} $$
Hence, the ratio of the radii is \( r_1 : r_2 = 5:4 \).

7. Find the angle in radian 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

Given: Pendulum length (radius) \( r = 75 \) cm.

To Find: Angle \( \theta \) in radians.

Formula: \( \theta = \frac{l}{r} \)

(i) Arc length \( l = 10 \) cm

Solution: $$ \theta = \frac{10}{75} = \frac{2}{15} \text{ radians} $$ The angle is \( \frac{2}{15} \) radians.

(ii) Arc length \( l = 15 \) cm

Solution: $$ \theta = \frac{15}{75} = \frac{1}{5} \text{ radians} $$ The angle is \( \frac{1}{5} \) radians.

(iii) Arc length \( l = 21 \) cm

Solution: $$ \theta = \frac{21}{75} = \frac{7}{25} \text{ radians} $$ The angle is \( \frac{7}{25} \) radians.