Trigonometric Functions

Class 11 NCERT

NCERT

1   Find the radian measures corresponding to the following degree measures:$\\$ (i) $25^o$$\\$ (ii)$-47^o30'$$\\$ (iii)$240^o$$\\$ (iv)$520^o$

Solution :

(i)$25^o$$\\$ We know that $180^o=\pi $ radian$\\$ $\therefore 25^o=\dfrac{\pi}{180}*25$ radian=$\dfrac{5 \pi}{36}$ radian$\\$ (ii)$-47^o30'$$\\$ $-47^o30'= -47\dfrac{1}{2}\\ =\dfrac{-95}{2}$degree$\\$ Since $180^o=\pi $ radian$\\$ $\dfrac{-95}{2}$degree$=\dfrac{\pi}{180}*(\dfrac{-95}{2})$ radian$\\$ $=(\dfrac{-19}{36*2})\pi$ radian$=\dfrac{-19}{72}\pi $ radian$\\$ $\therefore -47^o30'=\dfrac{-19}{72}\pi $ radian$\\$ (iii)$240^o$$\\$ We know that $180^o=\pi$ radian$\\$ $\therefore 240^o=\dfrac{\pi}{180}*240$ radian$\\$ $=\dfrac{4}{3}\pi$ radian$\\$ (iv) $ 520^o$$\\$ We know that $180^o=\pi$ radian$\\$ $\therefore 520^o=\dfrac{\pi}{180}*520$ radian$\\$ $=\dfrac{26\pi}{9}$ radian

2   Find the degree measures corresponding to the following radian measures(Use $\pi=\dfrac{22}{7}$)$\\$ (i)$\dfrac{11}{16}$$\\$ (ii)$-4$$\\$ (iii)$\dfrac{5\pi}{3}$$\\$ (iv)$\dfrac{7\pi}{6}$

Solution :

(i)$\dfrac{11}{16}$$\\$ We know that $\pi$radian$=180^o$$\\$ $\therefore \dfrac{11}{16}$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}$$\\$ We know that $\pi$radian$=180^o$$\\$ $\therefore \dfrac{7\pi}{6}$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 $\dfrac{20\pi}{3}cm$