# Trigonometric Functions

## Class 11 NCERT

### NCERT

##### Solution :

(i)$\dfrac{11}{16}$$\\ We know that \piradian=180^o$$\\$ $\therefore \dfrac{11}{16}$radian$=\dfrac{180}{\pi}*\dfrac{11}{16}$degree$\\$ $=\dfrac{45*11}{\pi*4}$degree$\\$ $=\dfrac{45*11*7}{22*4}$degree $\\$ $=\dfrac{315}{8}$degree$\\$ $39\dfrac{3}{8}$degree$\\$ $=39^o+\dfrac{3*60}{8}$minutes $\ \ \ \ \ \ \ \ \ \ [1^o=60']$$\\ =39^o+22'+\dfrac{1}{2}minutes\\ =39^o22'30'' \ \ \ \ \ \ \ \ \ \ \ \ [1'=60'']$$\\$ (ii)$-4$$\\ We know that \pi radian =180^o$$\\$ $-4$ radian =$\dfrac{180}{\pi}*(-4)$degree$\\$ $=\dfrac{180*7(-4)}{22}$degree$\\$ $=\dfrac{-2520}{11}$degree$=-229\dfrac{1}{11}$degree$\\$ $=-229^o-\dfrac{1*60}{11}$minutes $\ \ \ \ \ \ \ \ [1^o=60'] $$\\ =-229^o-5'-\dfrac{5}{11}minutes\\ =-229^o5'27'' \ \ \ \ \ \ \ \ \ [1'=60'']$$\\$ (iii)$\dfrac{5\pi}{3}$$\\ We know that \pi radian=180^o$$\\$ $\therefore \dfrac{5\pi}{3}$radian$\\$ $=\dfrac{180}{\pi}*\dfrac{5\pi}{3}$degree $\\$ $=300^o$ $\\$ (iv) $\dfrac{7\pi}{6}$$\\ We know that \piradian=180^o$$\\$ $\therefore \dfrac{7\pi}{6}$radian$\\$ $=\dfrac{180}{\pi}*\dfrac{7\pi}{6}\\ =210^o$

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

##### Solution :

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 $6 *2 \pi$ radian, i.e., $12 \pi$ radian$\\$ Thus, in one second, the wheel turns an angle of $12\pi$ radian

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$\pi =\dfrac{22}{7}$)

##### Solution :

We 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 =\dfrac{1}{r}$$\\ Therefore, for r =100 cm , l = 22 cm , we have\\ \theta =\dfrac{22}{100}radian =\dfrac{180}{\pi}*\dfrac{22}{100}degree\\ =\dfrac{180*7*22}{22*100}degree\\ =\dfrac{126}{10}degree\\ =12\dfrac{3}{5}degree=12^o36' \ \ \ \ \ \ \ [1^o=60']$$\\$ Thus, the required angle is $12^o 36'.$

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.

##### Solution :

Diameter of the circle = $40 cm$ $\therefore$ Radius $(r)$ of the circle =$\dfrac{40}{2} cm =20 cm$$\\ Let AB be a chord(length=20 cm) of the circle.\\ In \Delta OAB,OA=OB=\text{Radiusof circle}=20cm$$\\$ Also,$AB=20cm $$\\ Thus,\Delta OAB is an equilateral trigle.\\ \therefore \theta =60^o=\dfrac{\pi}{3} radian\\ We know that in a circle of radius r unit, if an arc of length l unit subtends an angle \theta radian at the centre then \\ \theta =\dfrac{l}{r}$$\\$ $\dfrac{\pi}{3}=\dfrac{\widehat{AB}}{20}\implies \widehat{AB}=\dfrac{20\pi}{3} cm$$\\$ Thus, the length of the minor arc of the chord is $\dfrac{20\pi}{3}cm$