Q1. Find the radian measures corresponding to the following degree measures:
(i) 25 ∘ 25^\circ 2 5 ∘
(ii) − 47 ∘ 30 ′ -47^\circ30' − 4 7 ∘ 3 0 ′
(iii) 240 ∘ 240^\circ 24 0 ∘
(iv) 520 ∘ 520^\circ 52 0 ∘
Sol. Radian measure = π 180 × =\frac{\pi}{180}\times = 180 π × Degree measure.
(i) 25 ∘ = π 180 × 25 = 5 π 36 radians
\begin{aligned}
\textbf{(i)}\ 25^\circ &=\frac{\pi}{180}\times25 \\[0.6em]
&=\frac{5\pi}{36}\text{ radians}
\end{aligned}
(i) 2 5 ∘ = 180 π × 25 = 36 5 π radians (ii) We know that 1 ∘ = 60 ′ 1^\circ=60' 1 ∘ = 6 0 ′ .
30 ′ = 30 60 ∘ = 1 2 ∘
30'=\frac{30}{60}^\circ=\frac12^\circ
3 0 ′ = 60 30 ∘ = 2 1 ∘ − 47 ∘ 30 ′ = − ( 47 + 1 2 ) ∘ = − ( 95 2 ) ∘
\begin{aligned}
-47^\circ30' &=-\left(47+\frac12\right)^\circ=-\left(\frac{95}{2}\right)^\circ
\end{aligned}
− 4 7 ∘ 3 0 ′ = − ( 47 + 2 1 ) ∘ = − ( 2 95 ) ∘ − ( 95 2 ) ∘ = π 180 × ( − 95 2 ) = − 19 π 72 radians
\begin{aligned}
-\left(\frac{95}{2}\right)^\circ &=\frac{\pi}{180}\times\left(-\frac{95}{2}\right) \\[0.6em]
&=-\frac{19\pi}{72}\text{ radians}
\end{aligned}
− ( 2 95 ) ∘ = 180 π × ( − 2 95 ) = − 72 19 π radians (iii) 240 ∘ = π 180 × 240 = 4 π 3 radians
\begin{aligned}
\textbf{(iii)}\ 240^\circ &=\frac{\pi}{180}\times240 \\[0.6em]
&=\frac{4\pi}{3}\text{ radians}
\end{aligned}
(iii) 24 0 ∘ = 180 π × 240 = 3 4 π radians (iv) 520 ∘ = π 180 × 520 = 26 π 9 radians
\begin{aligned}
\textbf{(iv)}\ 520^\circ &=\frac{\pi}{180}\times520 \\[0.6em]
&=\frac{26\pi}{9}\text{ radians}
\end{aligned}
(iv) 52 0 ∘ = 180 π × 520 = 9 26 π radians Q2. Find the degree measures corresponding to the following radian measures (Use π = 22 7 \pi=\frac{22}{7} π = 7 22 ).
(i) 11 16 \frac{11}{16} 16 11
(ii) − 4 -4 − 4
(iii) 5 π 3 \frac{5\pi}{3} 3 5 π
(iv) 7 π 6 \frac{7\pi}{6} 6 7 π
Sol. Degree measure = 180 π × =\frac{180}{\pi}\times = π 180 × Radian measure.
(i) Given, 11 16 \frac{11}{16} 16 11 radian. Using π = 22 7 \pi=\frac{22}{7} π = 7 22 ,
11 16 radian = ( 180 × 7 22 × 11 16 ) ∘ = ( 315 8 ) ∘ = 39 3 8 ∘
\begin{aligned}
\frac{11}{16}\text{ radian} &=\left(\frac{180\times7}{22}\times\frac{11}{16}\right)^\circ \\
&=\left(\frac{315}{8}\right)^\circ \\
&=39\frac38^\circ
\end{aligned}
16 11 radian = ( 22 180 × 7 × 16 11 ) ∘ = ( 8 315 ) ∘ = 39 8 3 ∘ We know that 1 ∘ = 60 ′ 1^\circ=60' 1 ∘ = 6 0 ′ and 1 ′ = 60 ′ ′ 1'=60'' 1 ′ = 6 0 ′′ .
= 39 ∘ + 3 8 × 60 ′ = 39 ∘ + 22 1 2 ′ = 39 ∘ + 22 ′ + 1 2 × 60 ′ ′ = 39 ∘ 22 ′ 30 ′ ′
\begin{array}{l}
\hspace{3em} \displaystyle =39^\circ+\frac38\times60' \\[0.8em]
\hspace{3em} \displaystyle =39^\circ+22\frac12' \\[0.8em]
\hspace{3em} \displaystyle =39^\circ+22'+\frac12\times60'' \\[0.8em]
\hspace{3em} \displaystyle =39^\circ22'30''
\end{array}
= 3 9 ∘ + 8 3 × 6 0 ′ = 3 9 ∘ + 22 2 1 ′ = 3 9 ∘ + 2 2 ′ + 2 1 × 6 0 ′′ = 3 9 ∘ 2 2 ′ 3 0 ′′ (ii) Given, − 4 -4 − 4 radians. Using π = 22 7 \pi=\frac{22}{7} π = 7 22 ,
− 4 radians = ( 180 × 7 22 × ( − 4 ) ) ∘ = − ( 2520 11 ) ∘ = − ( 229 + 1 11 ) ∘ = − ( 229 ∘ + 1 11 × 60 ′ ) = − ( 229 ∘ + 5 5 11 ′ ) = − ( 229 ∘ + 5 ′ + 5 11 × 60 ′ ′ ) = − ( 229 ∘ + 5 ′ + 27 3 11 ′ ′ ) ≈ − 229 ∘ 5 ′ 27 ′ ′
\begin{aligned}
-4\text{ radians} &=\left(\frac{180\times7}{22}\times(-4)\right)^\circ \\
&=-\left(\frac{2520}{11}\right)^\circ \\
&=-\left(229+\frac{1}{11}\right)^\circ \\
&=-\left(229^\circ+\frac{1}{11}\times60'\right) \\
&=-\left(229^\circ+5\frac{5}{11}'\right) \\
&=-\left(229^\circ+5'+\frac{5}{11}\times60''\right) \\
&=-\left(229^\circ+5'+27\frac{3}{11}''\right) \\
&\approx-229^\circ5'27''
\end{aligned}
− 4 radians = ( 22 180 × 7 × ( − 4 ) ) ∘ = − ( 11 2520 ) ∘ = − ( 229 + 11 1 ) ∘ = − ( 22 9 ∘ + 11 1 × 6 0 ′ ) = − ( 22 9 ∘ + 5 11 5 ′ ) = − ( 22 9 ∘ + 5 ′ + 11 5 × 6 0 ′′ ) = − ( 22 9 ∘ + 5 ′ + 27 11 3 ′′ ) ≈ − 22 9 ∘ 5 ′ 2 7 ′′ (iii) Given, 5 π 3 \frac{5\pi}{3} 3 5 π radians.
5 π 3 radians = ( 180 π × 5 π 3 ) ∘ = 300 ∘
\begin{aligned}
\frac{5\pi}{3}\text{ radians} &=\left(\frac{180}{\pi}\times\frac{5\pi}{3}\right)^\circ=300^\circ
\end{aligned}
3 5 π radians = ( π 180 × 3 5 π ) ∘ = 30 0 ∘ (iv) Given, 7 π 6 \frac{7\pi}{6} 6 7 π radians.
7 π 6 radians = ( 180 π × 7 π 6 ) ∘ = 210 ∘
\begin{aligned}
\frac{7\pi}{6}\text{ radians} &=\left(\frac{180}{\pi}\times\frac{7\pi}{6}\right)^\circ=210^\circ
\end{aligned}
6 7 π radians = ( π 180 × 6 7 π ) ∘ = 21 0 ∘ Q3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Sol. Number of revolutions in one minute = 360 =360 = 360 .
Number of revolutions in one second = 360 60 = 6 =\frac{360}{60}=6 = 60 360 = 6 .
Since 1 revolution = 360 ∘ = 2 π =360^\circ=2\pi = 36 0 ∘ = 2 π radians,
6 revolutions = 6 × 2 π radians = 12 π radians
\begin{array}{l}
\displaystyle 6\text{ revolutions}=6\times2\pi\text{ radians} \\[0.4em]
\hspace{5em} \displaystyle =12\pi\text{ radians}
\end{array}
6 revolutions = 6 × 2 π radians = 12 π radians ∴ Number of radians turned in one second = 12 π =12\pi = 12 π .
Q4. 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 = 22 7 \pi=\frac{22}{7} π = 7 22 ).
Sol. Given, arc length l = 22 cm l=22\text{ cm} l = 22 cm and radius r = 100 cm r=100\text{ cm} r = 100 cm .
We know that
θ = l r (radians) = 22 100 radians
\begin{aligned}
\theta &=\frac{l}{r}\text{ (radians)} \\
&=\frac{22}{100}\text{ radians}
\end{aligned}
θ = r l (radians) = 100 22 radians Degree measure = 180 π × Radian measure = 180 π × 22 100 = 180 × 7 22 × 22 100 = 12 3 5 ∘ = 12 ∘ + 3 5 × 60 ′ = 12 ∘ 36 ′
\begin{aligned}
\text{Degree measure} &=\frac{180}{\pi}\times\text{Radian measure} \\
&=\frac{180}{\pi}\times\frac{22}{100} \\
&=\frac{180\times7}{22}\times\frac{22}{100} \\
&=12\frac35^\circ \\
&=12^\circ+\frac35\times60' \\
&=12^\circ36'
\end{aligned}
Degree measure = π 180 × Radian measure = π 180 × 100 22 = 22 180 × 7 × 100 22 = 12 5 3 ∘ = 1 2 ∘ + 5 3 × 6 0 ′ = 1 2 ∘ 3 6 ′ Q5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Sol. Let O O O be the centre of the circle and A B AB A B be the chord.
Radius, r = 40 2 = 20 cm r=\frac{40}{2}=20\text{ cm} r = 2 40 = 20 cm , and A B = 20 cm AB=20\text{ cm} A B = 20 cm .
In △ O A B \triangle OAB △ O A B , O A = O B = A B = 20 cm OA=OB=AB=20\text{ cm} O A = O B = A B = 20 cm , so △ O A B \triangle OAB △ O A B is equilateral.
Hence,
∠ A O B = 60 ∘ = 60 × π 180 = π 3 radian
\angle AOB=60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radian}
∠ A O B = 6 0 ∘ = 60 × 180 π = 3 π radian Length of the minor arc,
l = r θ \textcolor{#1B5E20}{
l=r\theta
} l = r θ Substituting the values,
l = 20 × π 3 = 20 π 3 cm
\begin{array}{l}
\displaystyle l=20\times\frac{\pi}{3} \\[0.8em]
\hspace{0.6em} \displaystyle =\frac{20\pi}{3}\text{ cm}
\end{array}
l = 20 × 3 π = 3 20 π cm ∴ The length of the minor arc of the chord is 20 π 3 cm \frac{20\pi}{3}\text{ cm} 3 20 π cm .
Q6. If in two circles, arcs of the same length subtend angles 60 ∘ 60^\circ 6 0 ∘ and 75 ∘ 75^\circ 7 5 ∘ at the centre, find the ratio of their radii.
Sol. Let r 1 r_1 r 1 and r 2 r_2 r 2 be the radii of the two circles and l l l be the length of each arc.
Converting the angles into radians,
θ 1 = 60 ∘ = 60 × π 180 = π 3 radian θ 2 = 75 ∘ = 75 × π 180 = 5 π 12 radian
\begin{array}{l}
\displaystyle \theta_1=60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radian} \\[0.8em]
\displaystyle \theta_2=75^\circ=75\times\frac{\pi}{180}=\frac{5\pi}{12}\text{ radian}
\end{array}
θ 1 = 6 0 ∘ = 60 × 180 π = 3 π radian θ 2 = 7 5 ∘ = 75 × 180 π = 12 5 π radian We know that l = r θ l=r\theta l = r θ . Since the arcs are of the same length,
r 1 θ 1 = r 2 θ 2
r_1\theta_1=r_2\theta_2
r 1 θ 1 = r 2 θ 2 Substituting the values,
r 1 × π 3 = r 2 × 5 π 12 r 1 r 2 = 5 π 12 × 3 π r 1 r 2 = 5 4
\begin{array}{l}
\displaystyle r_1\times\frac{\pi}{3}=r_2\times\frac{5\pi}{12} \\[0.8em]
\displaystyle \frac{r_1}{r_2}=\frac{5\pi}{12}\times\frac{3}{\pi} \\[0.8em]
\displaystyle \frac{r_1}{r_2}=\frac54
\end{array}
r 1 × 3 π = r 2 × 12 5 π r 2 r 1 = 12 5 π × π 3 r 2 r 1 = 4 5 ∴ The ratio of their radii is r 1 : r 2 = 5 : 4 r_1:r_2=5:4 r 1 : r 2 = 5 : 4 .
Q7. 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
Sol. Here, the length of the pendulum is the radius, r = 75 cm r=75\text{ cm} r = 75 cm .
We know that
θ = l r \textcolor{#1B5E20}{
\theta=\frac{l}{r}
} θ = r l (i) Sol. For l = 10 cm l=10\text{ cm} l = 10 cm ,
θ = 10 75 = 2 15 radian
\theta=\frac{10}{75}=\frac{2}{15}\text{ radian}
θ = 75 10 = 15 2 radian ∴ The pendulum swings through 2 15 \frac{2}{15} 15 2 radian.
(ii) Sol. For l = 15 cm l=15\text{ cm} l = 15 cm ,
θ = 15 75 = 1 5 radian
\theta=\frac{15}{75}=\frac15\text{ radian}
θ = 75 15 = 5 1 radian ∴ The pendulum swings through 1 5 \frac15 5 1 radian.
(iii) Sol. For l = 21 cm l=21\text{ cm} l = 21 cm ,
θ = 21 75 = 7 25 radian
\theta=\frac{21}{75}=\frac{7}{25}\text{ radian}
θ = 75 21 = 25 7 radian ∴ The pendulum swings through 7 25 \frac{7}{25} 25 7 radian.