Exercise (10.1)-Trigonometry

Degree-Radian Relation

$\begin{aligned} \pi \text { radians } &=180 \text { degrees } \\\\ 1 \text { radian } \times \pi &=180 \text { degrees } \\\\ 1 \text { radian } &=\displaystyle\frac{180^{\circ}}{\pi} \\\\ &=57.2957 \text { degrees } \\\\ &\approx 57^{\circ} 18^{\prime} \end{aligned}$

$\begin{aligned} 180 \text { degree } &=\pi \text { radians } \\\\ 1 \text { degree } \times 180 &=\pi \text { radians } \\\\ 1 \text { degree } &=\displaystyle\frac{\pi}{180} \text { radians } \\\\ & \approx 0.01745 \text { radians } \end{aligned}$

Central Angle, Arc Length and Area of a Sector

(i) $\theta=\displaystyle\frac{s}{r}$

(ii) $s=r \theta$

(iii) $A=\displaystyle\frac{1}{2} r^{2} \theta$

where $\theta$ must be radian measure.

Exercise (10.1)- Problems and Solutions

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)	$390^{\circ}$






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



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 $A B D C .$
   





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

Exercise (1) - No.3 Solutions

(a) Angles (Degree) in Cartesian Plane

(b) Angles (Radian) in Cartesian Plane

 

Dynamic Representation

Exercise (1) - No.2 Solutions

$\displaystyle \displaystyle \begin{array}{{|l|}}\hline \displaystyle 1\ \text{radian} = \frac{{180{}^\circ }}{\pi } \\ \hline \end{array}$

$\displaystyle \displaystyle \begin{array}{{|l|}}\hline \displaystyle \theta\ \text{radians} =\theta\times \frac{{180{}^\circ }}{\pi } \\ \hline \end{array}$

$ \displaystyle \text{(a)}\ 2\pi \ \text{rad}=\text{2}\pi \times \frac{{180{}^\circ }}{\pi }=360{}^\circ $

$ \displaystyle \text{(b)}\ \frac{{11\pi }}{6}\ \text{rad}=\frac{{11\pi }}{6}\times \frac{{180{}^\circ }}{\pi }=330{}^\circ $

$ \displaystyle \text{(c)}\ \frac{{7\pi }}{4}\ \text{rad}=\frac{{7\pi }}{4}\times \frac{{180{}^\circ }}{\pi }=315{}^\circ $

$ \displaystyle \text{(d)}\ \frac{{5\pi }}{3}\ \text{rad}=\frac{{5\pi }}{3}\times \frac{{180{}^\circ }}{\pi }=330{}^\circ $

$ \displaystyle \text{(e) }\frac{{3\pi }}{2}\ \text{rad}=\frac{{3\pi }}{2}\times \frac{{180{}^\circ }}{\pi }=270{}^\circ $

$ \displaystyle \text{(f) }\frac{{4\pi }}{3}\ \text{rad}=\frac{{4\pi }}{3}\times \frac{{180{}^\circ }}{\pi }=240{}^\circ $

$ \displaystyle \text{(g) }\frac{{5\pi }}{4}\ \text{rad}=\frac{{5\pi }}{4}\times \frac{{180{}^\circ }}{\pi }=225{}^\circ $

$ \displaystyle \text{(h) }\frac{{7\pi }}{6}\ \text{rad}=\frac{{7\pi }}{6}\times \frac{{180{}^\circ }}{\pi }=210{}^\circ $

$ \displaystyle \text{(i) }\pi \ \text{rad}=\pi \times \frac{{180{}^\circ }}{\pi }=180{}^\circ $

$ \displaystyle \text{(j) }\frac{{5\pi }}{6}\ \text{rad}=\frac{{5\pi }}{6}\times \frac{{180{}^\circ }}{\pi }=150{}^\circ $

$ \displaystyle \text{(k) }\frac{{3\pi }}{4}\ \text{rad}=\frac{{3\pi }}{4}\times \frac{{180{}^\circ }}{\pi }=135{}^\circ $

$ \displaystyle \text{(l) }\frac{{2\pi }}{3}\ \text{rad}=\frac{{2\pi }}{3}\times \frac{{180{}^\circ }}{\pi }=120{}^\circ $

$ \displaystyle \text{(m) }\frac{\pi }{2}\ \text{rad}=\frac{\pi }{2}\times \frac{{180{}^\circ }}{\pi }=90{}^\circ $

$ \displaystyle \text{(n) }\frac{\pi }{3}\ \text{rad}=\frac{\pi }{3}\times \frac{{180{}^\circ }}{\pi }=60{}^\circ $

$ \displaystyle \text{(o) }\frac{\pi }{4}\ \text{rad}=\frac{\pi }{4}\times \frac{{180{}^\circ }}{\pi }=45{}^\circ $

$ \displaystyle \text{(p )}\frac{\pi }{6}\ \text{rad}=\frac{\pi }{6}\times \frac{{180{}^\circ }}{\pi }=30{}^\circ $
 

Exercise (1) - No.1 Solutions

$\displaystyle \displaystyle\ \begin{array}{{|l|}}\hline \displaystyle 1{}^\circ =\frac{\pi }{{180}}\ \text{radians} \\ \hline \end{array}$

$\displaystyle \displaystyle \ \begin{array}{{|l|}}\hline \displaystyle \theta{}^\circ =\theta\times \frac{\pi }{{180}}\ \text{radians} \\ \hline \end{array}$  

$ \displaystyle \text{(a)}\ 30{}^\circ =30\times \frac{\pi }{{180}}=\frac{\pi }{6}\ \text{rad}$

$ \displaystyle \text{(b)}\ 45{}^\circ =45\times \frac{\pi }{{180}}=\frac{\pi }{4}\ \text{rad}$

$ \displaystyle \text{(c)}\ 60{}^\circ =60\times \frac{\pi }{{180}}=\frac{\pi }{3}\ \text{rad}$

$ \displaystyle \text{(d)}\ 90{}^\circ =90\times \frac{\pi }{{180}}=\frac{\pi }{2}\ \text{rad}$

$ \displaystyle \text{(e)}\ 120{}^\circ =120\times \frac{\pi }{{180}}=\frac{{2\pi }}{3}\ \text{rad}$

$ \displaystyle \text{(f)}\ 135{}^\circ =135\times \frac{\pi }{{180}}=\frac{{3\pi }}{4}\ \text{rad}$

$ \displaystyle \text{(g)}\ 150{}^\circ =150\times \frac{\pi }{{180}}=\frac{{5\pi }}{6}\ \text{rad}$

$ \displaystyle \text{(h)}\ 180{}^\circ =180\times \frac{\pi }{{180}}=\pi \ \text{rad}$

$ \displaystyle \text{(i)}\ 210{}^\circ =210\times \frac{\pi }{{180}}=\frac{{7\pi }}{6}\ \text{rad}$

$ \displaystyle \text{(j)}\ 225{}^\circ =225\times \frac{\pi }{{180}}=\frac{{5\pi }}{4}\ \text{rad}$

$ \displaystyle \text{(k)}\ 240{}^\circ =240\times \frac{\pi }{{180}}=\frac{{4\pi }}{3}\ \text{rad}$

$ \displaystyle \text{(l)}\ 270{}^\circ =270\times \frac{\pi }{{180}}=\frac{{3\pi }}{2}\ \text{rad}$

$ \displaystyle \text{(m)}\ 300{}^\circ =300\times \frac{\pi }{{180}}=\frac{{5\pi }}{3}\ \text{rad}$

$ \displaystyle \text{(n)}\ 315{}^\circ =315\times \frac{\pi }{{180}}=\frac{{7\pi }}{4}\ \text{rad}$

$ \displaystyle \text{(o)}\ 330{}^\circ =330\times \frac{\pi }{{180}}=\frac{{11\pi }}{6}\ \text{rad}$

$ \displaystyle \text{(p)}\ 360{}^\circ =360\times \frac{\pi }{{180}}=2\pi \ \text{rad}$

Degree-Radian Converter

Degree မွ Radian သို႕ ေျပာင္းရာတြင္ Degree တန္ဖိုးေနရာတြင္ ႏွစ္သက္ရာ ကိန္းျပည့္ တန္ဖိုး ရိုက္ထည့္ပါ။ 

Radian  မွ Degree  သို႕ ေျပာင္းရာတြင္ ပိုင္းေ၀ တန္ဖိုးေနရာတြင္ ႏွစ္သက္ရာ ကိန္းျပည့္ တန္ဖိုး  ရိုက္ထည့္ပါ။ ပိုင္းေျခ တန္ဖိုးေနရာတြင္ ႏွစ္သက္ရာ အေပါင္းကိန္းျပည့္ တန္ဖိုး  ရိုက္ထည့္ပါ။ 

ထို႔ေနာက္ enter ေခါက္ပါ။ 

Relation between Degrees and Radians

The radian measure of an angle is the ratio of the length of the arc (s) whose centre is at the vertex of the angle to the radius (r).

$\displaystyle \begin{array}{{|l|}}\hline \displaystyle \theta =\frac{s}{r} \\ \hline \end{array}$

For one complete anticlockwise revolution, the length of the arc is the length of entire circumference and hence $ \displaystyle s=2\pi r$. Therefore, the radian measure of one complete revolution is

$\displaystyle \begin{array}{{|l|}}\hline \displaystyle \theta =\frac{2\pi r}{r}{}=2\pi \ \text{radians} \\ \hline \end{array}$  

In degrees, one complete counterclockwise revolution is $\displaystyle 360°$ and hence $ \displaystyle \theta=360°$.

$ \displaystyle \begin{array}{l}\therefore \ \ 360{}^\circ =2\pi \ \text{radians}\\\\\therefore \ \ 180{}^\circ =\pi \ \text{radians}\\\\\therefore \ \ 1{}^\circ \times 180=\pi \ \text{radians}\\\end{array}$

$\displaystyle \therefore\ $ $\displaystyle \begin{array}{{|l|}}\hline \displaystyle 1{}^\circ =\frac{\pi }{{180}}\ \text{radians} \\ \hline \end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{Since }\pi \ \text{radians}=180{}^\circ ,\\\\\ \ \ \ 1\ \text{radian}\ \times \pi =180{}^\circ \end{array}$

$\displaystyle \therefore\ $$\displaystyle \begin{array}{{|l|}}\hline \displaystyle 1\ \text{radian}\ =\frac{{180{}^\circ }}{\pi } \\ \hline \end{array}$