Drill for Exam

Exercise 10.1

1. Convert each of the following to radians.
   

   (a)	$120^{\circ}$	(b)	$90^{\circ}$
   (c)	$72^{\circ}$	(d)	$225^{\circ}$
   (e)	$150^{\circ}$	(f)	$108^{\circ}$
   (g)	$160^{\circ}$	(h)	$390^{\circ}$






$\begin{array}{ll} \text{(a)} & 120^{\circ}=120\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{2\pi}{3}\\\\ \text{(b)} & 90^{\circ}=90\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{\pi}{2}\\\\ \text{(c)} & 72^{\circ}=72\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{2\pi}{5}\\\\ \text{(d)} & 225^{\circ}=225\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{5\pi}{4}\\\\ \text{(e)} & 150^{\circ}=150\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{5\pi}{6}\\\\ \text{(f)} & 108^{\circ}=108\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{3\pi}{5}\\\\ \text{(g)} & 160^{\circ}=160\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{8\pi}{9}\\\\ \text{(h)} & 390^{\circ}=390\times \displaystyle\frac{\pi}{180}=\displaystyle\frac{13\pi}{6} \end{array}$



2. Convert each of the following to degrees.
   

   (a)	$\displaystyle\frac{\pi}{5}$	(b)	$\displaystyle\frac{3 \pi}{4}$
   (c)	$\displaystyle\frac{5 \pi}{6}$	(d)	$\pi$
   (e)	$\displaystyle\frac{8 \pi}{9}$	(f)	$\displaystyle\frac{12 \pi}{5}$
   (g)	$\displaystyle\frac{\pi}{3}$	(h)	$\displaystyle\frac{7 \pi}{3}$






$\begin{array}{ll} \text{(a)} & \displaystyle\frac{\pi}{5}=\displaystyle\frac{\pi}{5}\times \displaystyle\frac{180^{\circ}}{\pi}=36^{\circ}\\\\ \text{(b)} & \displaystyle\frac{3\pi}{4}=\displaystyle\frac{3\pi}{4}\times \displaystyle\frac{180^{\circ}}{\pi}=135^{\circ}\\\\ \text{(c)} & \displaystyle\frac{5\pi}{6}=\displaystyle\frac{5\pi}{6}\times \displaystyle\frac{180^{\circ}}{\pi}=150^{\circ}\\\\ \text{(d)} & \pi=\pi\times \displaystyle\frac{180^{\circ}}{\pi}=180^{\circ}\\\\ \text{(e)} & \displaystyle\frac{8\pi}{9}=\displaystyle\frac{8\pi}{9}\times \displaystyle\frac{180^{\circ}}{\pi}=160^{\circ}\\\\ \text{(f)} & \displaystyle\frac{12\pi}{5}=\displaystyle\frac{12\pi}{5}\times \displaystyle\frac{180^{\circ}}{\pi}=432^{\circ}\\\\ \text{(g)} & \displaystyle\frac{\pi}{3}=\displaystyle\frac{\pi}{3}\times \displaystyle\frac{180^{\circ}}{\pi}=60^{\circ}\\\\ \text{(h)} & \displaystyle\frac{7\pi}{3}=\displaystyle\frac{7\pi}{3}\times \displaystyle\frac{180^{\circ}}{\pi}=210^{\circ}\\\\ \end{array}$



3. A central angle $\theta$ subtends an arc of $\displaystyle\frac{11 \pi}{2}$ cm on a circle of radius $6$ cm. Find the measure of $\theta$ in radians and the area of a sector of a circle which has $\theta$ is its central angle.





$\begin{array}{l}\text{arc length}\ =s=\displaystyle\frac{{11\pi }}{2}\ \ \text{cm}\\\\\text{radius}\ =r=\ 6\ \text{cm}\ \\\\\text{centeral angle}\ =\theta =?\\\\\text{area of sector}\ =A=?\\\\\theta =\displaystyle\frac{s}{r}\,=\displaystyle\frac{{\displaystyle\frac{{11\pi }}{2}}}{6}=\displaystyle\frac{{11\pi }}{{12}}\ \text{radians}\\\\A=\displaystyle\frac{1}{2}{{r}^{2}}\theta \\\\\ \ \ =\displaystyle\frac{1}{2}\times {{6}^{2}}\times \displaystyle\frac{{11\pi }}{{12}}\\\\\ \ =\displaystyle\frac{{33}}{2}=16.5\ \text{c}{{\text{m}}^{2}}\end{array}$



4. The area of a sector of a circle is $143 \mathrm{~cm}^{2}$ and the length of the arc of a sector is $11\ \mathrm{~cm}$. Find the radius of the circle.





$\begin{array}{l}\text{Area}\ =A=143\ \ \text{c}{{\text{m}}^{2}}\\\\\text{arc length}\ =s=\ 11\ \text{cm}\ \\\\\text{radius}\ =r=?\ \\\\A=\displaystyle\frac{1}{2}{{r}^{2}}\left( {\displaystyle\frac{s}{r}\,} \right)\ \ \left( {\because \theta =\displaystyle\frac{s}{r}} \right)\\\\143\ =\displaystyle\frac{1}{2}\times r\times 11\\\\r=26\ \text{cm}\end{array}$



5. A sector cut from a circle of radius $3 \mathrm{~cm}$ has a perimeter of $16 \mathrm{~cm}$. Find the area of this sector.





$\begin{array}{l}\text{radius}\ =r=3\ \ \text{cm}\\\\\text{perimeter}\ =s+2r=\ 16\ \text{cm}\ \\\\\therefore \ \ s=16-6=10\ \text{cm}\\\\\text{Area}\ =A=?\ \\\\\text{Since }\theta =\displaystyle\frac{s}{r},\ s=r\theta \\\\A=\displaystyle\frac{1}{2}{{r}^{2}}\theta \ \ \\\\\ \ \ =\displaystyle\frac{1}{2}r(r\theta )\\\\\ \ \ =\displaystyle\frac{1}{2}rs\\\\\ \ \ =\displaystyle\frac{1}{2}\times 3\times \ 10\ \ \ \ \ \\\\A=15\ \text{c}{{\text{m}}^{2}}\end{array}$



6. A piece of wire of fixed length $L\ \mathrm{~cm}$, is bent to form the boundary a sector of a circle. The circle has radius $r\ \mathrm{~cm}$ and the angle of the sector is $\theta=\left(\displaystyle\frac{32}{r}-2\right)$ radians. Find the wire of fixed length $L$ and show that the area of the sector, $A \mathrm{~cm}^{2}$ is given by $A=16 r-r^{2}$.





$\begin{array}{l}\text{radius}\ =r=3\ \ \text{cm}\\\\\text{perimeter}\ =s+2r=\ 16\ \text{cm}\ \\\\\therefore \ \ s=16-6=10\ \text{cm}\\\\\text{Since the wire is bent into a sector of circle,}\\\\\text{perimeter}=L\ \text{cm}\ \\\\\therefore \ \ s+2r=L\\\\\text{radius}=r\ \text{cm}\\\\\text{central angle}=\theta =\left( {\displaystyle\frac{{32}}{r}-2} \right)\ \text{radians}\ \\\\\text{Since }\theta =\displaystyle\frac{s}{r},\ s=r\theta \\\\\therefore L=s+2r\\\\\ \ \ \ \ \ =r\theta +2r\\\\\ \ \ \ \ \ =r\left( {\theta +2} \right)\\\\\ \ \ \ \ \ =r\left( {\displaystyle\frac{{32}}{r}-2+2} \right)\\\\\ \ \ \ \ \ =32\ \text{cm}\\\\A\ \ \ =\displaystyle\frac{1}{2}{{r}^{2}}\theta \\\\\ \ \ =\displaystyle\frac{1}{2}\times {{r}^{2}}\times \ \left( {\displaystyle\frac{{32}}{r}-2} \right)\ \ \ \ \ \\\\\therefore A=16r-{{r}^{2}}\end{array}$



7. A race is run at a uniform speed on a circular course. In each minute, a runner traverses an arc of a circle which subtends $2 \displaystyle\frac{6}{7}$ radians at the centre of the course. If each lap is $792$ yards, how long does the runner take to run a mile?





$\begin{array}{l}\text{The central angle of the arc for which }\\\text{the runner take in each minute }=\theta =2\displaystyle\frac{6}{7}=\displaystyle\frac{{20}}{7}\ \text{radians}\ \\\\\text{The length of each lap}\ =\ \text{circumference}=792\ \ \text{yards}\\\\\therefore \ \ 2\pi r=\ 792\ \text{yards}\ \\\\\therefore \ \ r=\ \displaystyle\frac{{792}}{{2\pi }}\ \text{yards}\\\\\text{Let the distance travelled by runner in each minute }=s\\\\\text{Since}\ \theta =\displaystyle\frac{s}{r},\ s=r\theta \\\\\therefore \ \ s=\displaystyle\frac{{792}}{{2\pi }}\times \displaystyle\frac{{20}}{7}=\displaystyle\frac{{7920}}{{7\pi }}\ \text{yards}\\\\\therefore \ \ \text{speed of runner}=\displaystyle\frac{{7920}}{{7\pi }}\ \text{yards}/\text{min}\\\\\text{Let the time taken by runner to run 1 mile (1760 yards)}=t\ \text{min}\ \\\\\therefore \ \ \displaystyle\frac{{1760}}{t}=\displaystyle\frac{{7920}}{{7\pi }}\\\\\therefore \ \ t=1760\times \displaystyle\frac{{7\pi }}{{7920}}\\\\\therefore \ \ t=4.8869\ \text{min}\ \end{array}$



8. The large hand of a clock is $28$ inches long; how many inches does its extremity move in $20$ minutes?





$\begin{array}{l}\text{The length of the hand}=r=28\text{ in }\\\\\text{Let}\ \text{the angle taken by the hand in 20 minutes }=\theta \\\\\therefore \ \ \theta =\displaystyle\frac{{360}}{{60}}\times 20\times \displaystyle\frac{\pi }{{180}}=\displaystyle\frac{{2\pi }}{3}\ \text{radians}\\\\\text{Let the arc length taken by the hand}\ =\ s\\\\\text{Since}\ \theta =\displaystyle\frac{s}{r},\ s=r\theta \\\\\therefore \ \ s=28\times \displaystyle\frac{{2\pi }}{3}=\displaystyle\frac{{56\pi }}{3}\ \text{in}=\ 58.64\ \text{in}\end{array}$



9. The figure shows two sectors in which the arcs $A B$ and $C D$ are arcs of concentric circles, centre $O$. If $\angle A O B=\displaystyle\frac{2}{3}$ radians, $A C=3 \mathrm{~cm}$ and the area of a sector $A O B$ is $12 \mathrm{~cm}^{2}$, calculate the area and the perimeter of $A B D C .$
   





$\begin{array}{l}\text{Let}\ \angle AOB\text{ }=\theta \\\\\therefore \ \ \theta =\displaystyle\frac{2}{3}\ \text{radian}\\\\\text{Let the}\ \text{area of sector }AOB\ =\ {{A}_{1}}\\\\\therefore \ {{A}_{1}}=12\ \text{c}{{\text{m}}^{2}}\\\\\text{Let }OA\ =\ {{r}_{1}}\ \text{and}\ OC\ =\ {{r}_{2}}\\\\\therefore \ \ {{r}_{2}}={{r}_{1}}+3\\\\{{A}_{1}}=12\\\\\displaystyle\frac{1}{2}{{r}_{1}}^{2}\ \theta =12\\\\\displaystyle\frac{1}{2}{{r}_{1}}^{2}\ \left( {\displaystyle\frac{2}{3}} \right)=12\\\\\therefore \ \ {{r}_{1}}^{2}=36\Rightarrow {{r}_{1}}=6\ \text{cm}\\\\\therefore \ \ {{r}_{2}}={{r}_{1}}+3=6+3=9\ \text{cm}\\\\\text{Let the}\ \text{area of sector }COD\ =\ {{A}_{2}}\\\\\therefore \ {{A}_{2}}=\displaystyle\frac{1}{2}{{r}_{2}}^{2}\ \theta =\displaystyle\frac{1}{2}\left( {{{9}^{2}}} \right)\ \left( {\displaystyle\frac{2}{3}} \right)=27\ \text{c}{{\text{m}}^{2}}\\\\\therefore \ \ \text{the}\ \text{area of}\ ABDC\\\\=\ {{A}_{2}}-{{A}_{1}}\\\\=27-12\\\\=15\ \ \text{c}{{\text{m}}^{2}}\\\\\text{Let the}\ \text{length of}\ \text{arc }AB\ \text{be}\ {{s}_{1}}\text{ and that of }CD\ \text{be}\ {{s}_{2}}.\\\\\text{Since }\theta =\displaystyle\frac{{{{s}_{1}}}}{{{{r}_{1}}}}=\displaystyle\frac{{{{s}_{2}}}}{{{{r}_{2}}}},\\\\{{s}_{1}}={{r}_{1}}\theta =6\left( {\displaystyle\frac{2}{3}} \right)=4\ \text{cm}\\\\{{s}_{2}}={{r}_{2}}\theta =9\left( {\displaystyle\frac{2}{3}} \right)=6\ \text{cm}\\\\\text{the}\ \text{perimeter of}\ ABDC={{s}_{1}}+{{s}_{2}}+3+3\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 4+6+3+3\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 16\ \text{cm}\end{array}$