Chapter 3: Trigonometric Functions

Ans: As we all know that ${180^\circ } = \pi $ radian. So, ${25^\circ } = \frac{\pi }{{180}} \times 25$ radian $ = \frac{{5\pi }}{{36}}$ radian.

(ii) -47o30’

Ans: As we all know that, $ - {47^\circ }{30^\prime } = \,\, - 47\frac{1}{2}$

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

Because of the ${180^\circ } = \pi $ radian $\frac{{ - 95}}{2}$ degree $ = \frac{\pi }{{180}} \times \left( {\frac{{ - 95}}{2}} \right)\operatorname{radian}  = \left( {\frac{{ - 19}}{{36 \times 2}}} \right)\pi $ radian $ = \frac{{ - 19}}{{72}}\pi $ radian.

So, $ - {47^\circ }{30^\prime } = \frac{{ - 19}}{{72}}\pi $ radian.

(iii) 240o

Ans: As we all know that ${180^\circ } = \pi $ radian

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

(iv) 520o

Ans: As we all know that ${180^\circ } = \pi $ radian. So, ${520^\circ } = \frac{\pi }{{180}} \times 520$ radian $ = \,\frac{{26\pi }}{9}$radian.

Ans: As we all know that $\pi $ radian $ = {180^\circ }$. So, $\frac{{11}}{{16}}$ radian $ = \frac{{180}}{\pi } \times \frac{{11}}{{16}}$ degree $ = \frac{{45 \times 11}}{{\pi  \times 4}}$ degree

$ = \frac{{45 \times 11 \times 7}}{{22 \times 4}}$ degree $ = \frac{{315}}{8}$ degree

$ = 36\frac{3}{8}$ degree

$ = {39^\circ } + \frac{{3 \times 60}}{8}$ minutes $\quad \left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\circ } = {{60}^\prime }} \right]$

$ = {39^\circ } + {22^\prime } + \frac{1}{2}$ minutes

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

(ii) -4

Ans: As we all know that $\pi $ radian $ = {180^\circ }$. So, $ - 4$ radian $ = \frac{{180}}{\pi } \times ( - 4)$ degree $ = \frac{{180 \times 7( - 4)}}{{22}}$ degree

$ = \frac{{ - 2520}}{{11}}$ degree $ =  - 229\frac{1}{{11}}$ degree

$ =  - {229^\circ } + \frac{{1 \times 60}}{{11}}$ minutes

$\left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\circ } = {{60}^\prime }} \right]$

$ =  - {229^\circ } + {5^\prime } + \frac{5}{{11}}$ minutes

$ =  - {229^\circ }{5^\prime }{27^{\prime \prime }}\quad \left[ {{\text{We}}\,\,{\text{know}}\,\,{\text{that}}\,\,{1^\prime } = {{60}^{\prime \prime }}} \right]$

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

Ans: As we all know that $\pi $ radian$ = {180^\circ }$. So,\[\frac{{5\pi }}{3}\] radian $ = \frac{{180}}{\pi } \times \frac{{5\pi }}{3}$ degree $ = {300^\circ }$

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

Ans: As we all know that $\pi $ radian$ = {180^\circ }$. So, $\frac{{7\pi }}{6}$ radian $ = \frac{{180}}{\pi } \times \frac{{7\pi }}{6} = {210^\circ }$.

Ans: The number of revolutions mades by the wheel in one minute$ = 360$

As a result, the number of revolutions made by the wheel in one second$ = \frac{{360}}{{60}} = 6$ 

The wheel rotates at a $2\pi $radian angle in one complete revolution. As a result, it will turn an angle of $6 \times 2\pi $radian in 6 complete revolutions, i.e., $12\pi $ radian

As a result, the wheel rotates at a $12\pi $radian angle in one second.

Ans: As we all know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta $ radian at the centre, then $\theta  = \frac{1}{r}$

Hence, for $r = 100\;{\text{cm}},l = 22\;{\text{cm}}$, we have $\theta  = \frac{{22}}{{100}}$ radian $ = \frac{{180}}{\pi } \times \frac{{22}}{{100}}$ degree $ = \frac{{180 \times 7 \times 22}}{{22 \times 100}}$ degree

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

As a result, the needed angle is ${12^\circ }{36^\prime }$.

Ans: The diameter of the circle $ = 40\;{\text{cm}}$

So, Radius $(r)$ of the circle $ = \frac{{40}}{2}\;{\text{cm}} = 20\;{\text{cm}}$

Let ${\text{AB}}$ be a chord ( length $ = 20\;{\text{cm}})$ of the circle.

In $\Delta OAB,OA = OB = $ Radius of circle $ = 20\;{\text{cm}}$

Also, ${\text{AB}} = 20\;{\text{cm}}$

As a result, $\Delta OAB$ is an equilateral triangle.

So, $\theta  = {60^\circ } = \frac{\pi }{3}$ radian

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

$\frac{\pi }{3} = \frac{{\widehat {AB}}}{{20}} \Rightarrow \widehat {AB} = \frac{{20\pi }}{3}\;{\text{cm}}$

As a result, the length of the minor arc of the chord is $\frac{{20\pi }}{3}\,{\text{cm}}.$

Ans: The radii of the two circles should be ${r_1}$ and ${r_2}$. Whereas an arc of length $l$ 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, ${60^\circ } = \frac{\pi }{3}$ radian and ${75^\circ } = \frac{{5\pi }}{{12}}$ radian. We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends an angle $\theta $. radian at the centre then $\theta  = \frac{l}{r}$ or $l = r\theta $

So, $l = \frac{{{r_1}\pi }}{3}$ and $l = \frac{{{r_2}5\pi }}{{12}}$

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

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

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

As a result, the ratio of the radii is 5: 4.

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta $ radian at the centre, then $\theta  = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

Now, $l = 10\;{\text{cm}}$

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

(ii) 15cm

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta $ radian at the centre, then $\theta  = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

 Now, $l = 15\;{\text{cm}}$

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

(iii) 21cm

Ans: We also know that in a circle of radius $r$ unit, if an arc of length $l$ unit subtends An angle $\theta $ radian at the centre, then $\theta  = \frac{l}{r}$ It is given that $r = 75\;{\text{cm}}$.

Now, $l = 21\;{\text{cm}}$

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