NCERT Solutions for Class 11 Maths Chapter 3 - Trigonometric Functions Exercise 3.1

Exercise 3.1

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

(i) 25o

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.

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

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

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 }$.

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

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.

4: Find the degree measure of the angle subtended at the centre of a circle of radius $100\;cm$by an arc of length $22\;cm$. $\left( {} \right.$ Use $\left. {\pi  = \frac{{22}}{7}} \right)$

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 ${12^\circ }{36^\prime }$.

5: In a circle of diameter$40\;cm$, the length of a chord is $20\;cm$. Find the length of minor arc of the chord.

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 $\frac{{20\pi }}{3}\,{\text{cm}}.$

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.

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.

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) 10cm

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

Conclusion

Class 11 Maths Chapter 3 Exercise 3.1 Maths is essential for understanding the basics of trigonometric functions. It covers key concepts like angles, degree and radian measures, and their conversions. Focus on mastering these conversions and the application of radian measures in calculating arc lengths and sector areas. Vedantu provides the best solutions, offering clear explanations and step-by-step guidance. By regularly practising with these expert solutions of ex 3.1 class 11, students can strengthen their problem-solving skills and prepare well for their exams.

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Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.