Answer :

(i) $25^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 25^{\circ}=\dfrac{\pi}{180} \times 25$ radian $=\dfrac{5 \pi}{36}$ radian

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

$- 47^{\circ} 30^{\prime}={ }^{-47 \dfrac{1}{2}}$ degree $[1^{\circ}=60^{\prime}]$

$=\dfrac{-95}{2}$ degree

Since $180^{\circ}=\pi$ radian

$\dfrac{-95}{2}$ deg ree $=\dfrac{\pi}{180} \times(\dfrac{-95}{2})$ radian $=(\dfrac{-19}{36 \times 2}) \pi$ radian $=\dfrac{-19}{72} \pi$ radian

$\therefore-47^{\circ} 30^{\prime}=\dfrac{-19}{72} \pi$ radian

(iii) $240^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 240^{\circ}=\dfrac{\pi}{180} \times 240$ radian $=\dfrac{4}{3} \pi$ radian

(iv) $520^{\circ}$

We know that $180^{\circ}=\pi$ radian

$\therefore 520^{\circ}=\dfrac{\pi}{180} \times 520$ radian $=\dfrac{26 \pi}{9}$ radian

Answer :

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

We know that $\pi$ radian $=180^{\circ}$

$\therefore \dfrac{11}{16}$ radain $=\dfrac{180}{\pi} \times \dfrac{11}{16}$ deg ree $=\dfrac{45 \times 11}{\pi \times 4}$ deg ree

$=\dfrac{45 \times 11 \times 7}{22 \times 4}$ deg ree $=\dfrac{315}{8}$ deg ree

$=39 \dfrac{3}{8}$ deg ree

$=39^{\circ}+\dfrac{3 \times 60}{8}$ min utes $\quad[1^{\circ}=60^{\prime}]$

$=39^{\circ}+22^{\prime}+\dfrac{1}{2}$ min utes

$=39^{\circ} 22^{\prime} 30^{\prime \prime} \quad[1^{\prime}=60^{\prime \prime}]$

(ii) -4

We know that $\pi$ radian $=180^{\circ}$

$ \begin{aligned} -4 \text{ radian } & =\dfrac{180}{\pi} \times(-4) \text{ deg ree }=\dfrac{180 \times 7(-4)}{22} \text{ deg ree } \\ & =\dfrac{-2520}{11} \text{ deg ree }=-229 \dfrac{1}{11} \text{ deg ree } \\ & =-229^{\circ}+\dfrac{1 \times 60}{11} \text{ min utes } \quad[1^{\circ}=60^{\prime}] \\ & =-229^{\circ}+5^{\prime}+\dfrac{5}{11} \text{ min utes } \\ & =-229^{\circ} 5^{\prime} 27^{\prime \prime} \quad[1^{\prime}=60^{\prime \prime}] \end{aligned} $

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

We know that $\pi$ radian $=180^{\circ}$ $\therefore \dfrac{5 \pi}{3}$ radian $=\dfrac{180}{\pi} \times \dfrac{5 \pi}{3}$ deg ree $=300^{\circ}$

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

We know that $\pi$ radian $=180^{\circ}$

$\therefore \dfrac{7 \pi}{6}$ radian $=\dfrac{180}{\pi} \times \dfrac{7 \pi}{6}=210^{\circ}$

Answer :

Number of revolutions made by the wheel in 1 minute $=360$

$\therefore$ Number of revolutions made by the wheel in 1 second $=\dfrac{360}{60}=6$

In one complete revolution, the wheel turns an angle of $2 \pi$ radian.

Hence, in 6 complete revolutions, it will turn an angle of $6 \times 2 \pi$ radian, i.e.,

$12 \pi$ radian

Thus, in one second, the wheel turns an angle of $12 \pi$ radian.

Answer :

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\dfrac{1}{r}$

Therefore, r $=100 cm, I=22 cm$, we have

$\theta=\dfrac{22}{100}$ radian $=\dfrac{180}{\pi} \times \dfrac{22}{100}$ deg ree $=\dfrac{180 \times 7 \times 22}{22 \times 100}$ degree

$=\dfrac{126}{10}$ degree $=12 \dfrac{3}{5}$ degree $=12^{\circ} 36^{\prime} \quad[1^{\circ}=60^{\prime}]$

Thus, the required angle is $12^{\circ} 36^{’}$

Answer :

Diameter of the circle $=40 cm$

$\therefore$ Radius $(r)$ of the circle $=\dfrac{40}{2} cm=20 cm$

Let $A B$ be a chord (length $=20 cm$ ) of the circle.

In $\triangle OAB, OA=OB=$ Radius of circle $=20 cm$

Also, $A B=20 cm$

Thus, $\triangle OAB$ is an equilateral triangle.

$\therefore \theta=60^{\circ}=\dfrac{\pi}{3}$ radian

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\dfrac{l}{r}$

$\dfrac{\pi}{3}=\dfrac{\overline{AB}}{20} \rightarrow \overline{AB}=\dfrac{20 \pi}{3} cm$

Answer :

Let the radii of the two circles be ${ }^{r_1}$ and ${ }^{r_2}$. Let an arc of length / subtend an angle of $60^{\circ}$ at the centre of the circle of radius $r_1$, while let an arc of length / subtend an angle of $75^{\circ}$ at the centre of the circle of radius $r_2$.

Now, $60^{\circ}=\dfrac{\pi}{3}$ radian and $75^{\circ}=\dfrac{5 \pi}{12}$ radian

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then

$\theta=\dfrac{l}{r}$ or $l=r \theta$

$\therefore l=\dfrac{r_1 \pi}{3}$ and $l=\dfrac{r_2 5 \pi}{12}$

$\Rightarrow \dfrac{r_1 \pi}{3}=\dfrac{r_2 5 \pi}{12}$

$\Rightarrow r_1=\dfrac{r_2 5}{4}$

$\Rightarrow \dfrac{r_1}{r_2}=\dfrac{5}{4}$

Thus, the ratio of the radii is 5:4.

Answer :

We know that in a circle of radius $r$ unit, if an arc of length / unit subtends an angle $\theta$ radian at the centre, then $\theta=\dfrac{l}{r}$

It is given that $r=75 cm$

(i) Here, $I=10 cm$

$\theta=\dfrac{10}{75}$ radian $=\dfrac{2}{15}$ radian

(ii) Here, $I=15 cm$

$\theta=\dfrac{15}{75}$ radian $=\dfrac{1}{5}$ radian

(iii) Here, $I=21 cm$

$\theta=\dfrac{21}{75}$ radian $=\dfrac{7}{25}$ radian