Miscellaneous Exercise : Proofs of Trigonometric Identities

1. Prove that $\dfrac{\tan \theta+\cot \theta}{\sec \theta+\operatorname{cosec} \theta}=\dfrac{1}{\sin \theta+\cos \theta}$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\sec \theta+\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\cos \theta}+\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \dfrac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}\\\\ = &\dfrac{1}{\sin \theta+ \cos \theta} \end{aligned}$



2. Prove that $\quad \sec x \csc x-\cot x=\tan x$.





$\begin{aligned} & \sec x \csc x-\cot x \\\\ = & \dfrac{1}{\cos x} \dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x} \\\\ = & \dfrac{1-\cos ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin ^{2} x}{\sin x \cos x} \\\\ = & \dfrac{\sin x}{\cos x} \\\\ = & \tan x \end{aligned}$



3. Prove that $\dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x}=2 \csc^{2} x$





$\begin{aligned} & \dfrac{1}{1-\cos x}+\dfrac{1}{1+\cos x} \\\\ = & \dfrac{1+\cos x+1-\cos x}{(1-\cos x)(1+\cos x)} \\\\ = & \dfrac{2}{1-\cos ^{2} x} \\\\ = & \dfrac{2}{\sin ^{2} x} \\\\ = & 2 \csc ^{2} x \end{aligned}$



4. Show that $\dfrac{\tan \theta+\cot \theta}{\csc \theta}=\sec \theta$





$\begin{aligned} &\dfrac{\tan \theta+\cot \theta}{\csc \theta}\\\\ = &\dfrac{\dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{\dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}}{\dfrac{1}{\sin \theta}}\\\\ = &\dfrac{1}{\sin \theta \cos \theta} \times \sin \theta \\\\ = &\sec \theta \end{aligned}$



5. Show that $\sqrt{\sec ^{2} \theta-1}+\sqrt{\csc^{2} \theta-1}=\sec \theta \csc \theta$





$\begin{aligned} & \sqrt{\sec ^{2} \theta-1}+\sqrt{\csc ^{2} \theta-1} \\\\ =& \sqrt{\tan ^{2} \theta}+\sqrt{\cot ^{2} \theta} \\\\ =& \tan \theta+\cot \theta \\\\ =& \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta} \\\\ =& \dfrac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta} \\\\ =& \dfrac{1}{\sin \theta \cos \theta} \\\\ =& \sec \theta \csc \theta \end{aligned}$



6. Prove that $\sec ^{2} x+\csc^{2} x=\sec ^{2} x \csc^{2} x$





$\begin{aligned} & \sec ^{2} x+\csc ^{2} x \\\\ =& \dfrac{1}{\cos ^{2} x}+\dfrac{1}{\sin ^{2} x} \\\\ =& \dfrac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\sin ^{2} x \cos ^{2} x} \\\\ =& \dfrac{1}{\cos ^{2} x} \cdot \dfrac{1}{\sin ^{2} x}\\\\ =&\sec ^{2} x \csc ^{2} x \end{aligned}$



7. Show that $\dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1}=2 \tan ^{2} \theta$





$\begin{aligned} & \dfrac{1}{\csc \theta-1}-\dfrac{1}{\csc \theta+1} \\\\ =& \dfrac{(\csc \theta+1)-(\csc \theta-1)}{(\csc \theta-1)(\csc \theta+1)} \\\\ =& \dfrac{2}{\csc ^{2} \theta-1} \\\\ =& \dfrac{2}{\cot ^{2} \theta} \\\\ =& 2 \tan ^{2} \theta \end{aligned}$



8. Show that $\dfrac{\csc \theta}{\csc \theta-\sin \theta}=\sec ^{2} \theta$.





$\begin{aligned} &\dfrac{\csc \theta}{\csc \theta-\sin \theta} \\\\ =& \dfrac{\dfrac{1}{\sin \theta}}{\dfrac{1}{\sin \theta}-\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \times \dfrac{\sin \theta}{\cos ^{2} \theta}\\\\ =&\sec ^{2} \theta \end{aligned}$



9. Prove that $\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x}=\cos x-\sin x$.





$\begin{aligned} &\dfrac{\cos x}{1+\tan x}-\dfrac{\sin x}{1+\cot x} \\\\ =&\dfrac{\cos x}{1+\dfrac{\sin x}{\cos x}}-\dfrac{\sin x}{1+\dfrac{\cos x}{\sin x}} \\\\ =&\dfrac{\cos x+\sin x}{\cos x}-\dfrac{\sin x}{\dfrac{\sin x+\cos u}{\sin x}}\\\\ =&\cos x\left(\dfrac{\cos x}{\cos x+\sin x}\right)-\sin x\left(\dfrac{\sin x}{\sin x+\cos x}\right)\\\\ =&\dfrac{\cos ^{2} x-\sin ^{2} x}{\cos x+\sin x}\\\\ =&\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x+\sin x}\\\\ =&\cos x-\sin x \end{aligned}$



10. Show that $\csc \theta-\sin \theta=\cot \theta \cos \theta$.





$\begin{aligned} & \csc \theta-\sin \theta \\\\ =& \dfrac{1}{\sin \theta}-\sin \theta \\\\ =& \dfrac{1-\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{\cos \theta}{\sin \theta} \cos \theta \\\\ =& \cot \theta \cos \theta \end{aligned}$



11. Show that $\dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta}=\tan \theta \sin \theta$.





$\begin{aligned} & \dfrac{\tan ^{2} \theta+\sin ^{2} \theta}{\cos \theta+\sec \theta} \\\\ =& \dfrac{\dfrac{\sin ^{2} \theta}{\cos ^{2} \theta}+\sin ^{2} \theta}{\cos \theta+\dfrac{1}{\cos \theta}} \\\\ =& \dfrac{\sin ^{2} \theta\left(1+\cos ^{2} \theta\right)}{\cos ^{2} \theta} \times \dfrac{\cos \theta}{\cos ^{2} \theta+T}\\\\ =&\dfrac{\sin ^{2} \theta}{\cos \theta} \\\\ =&\dfrac{\sin \theta}{\cos \theta} \sin \theta \\\\ =&\tan \theta \sin \theta \end{aligned}$



12. Prove that $\sin x(\cot x+\tan x)=\sec x$





$\begin{aligned} & \sin x(\cot x+\tan x) \\ =& \sin x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right) \\\\ =& \sin x\left(\dfrac{\cos 2 x+\sin ^{2} x}{\sin x \cos x}\right) \\\\ =& \dfrac{1}{\cos x} \\\\ =& \sec x \end{aligned}$



13. Show that $\dfrac{\sec \theta}{\cot \theta+\tan \theta}=\sin \theta$.





$\begin{aligned} &\dfrac{\sec \theta}{\cot \theta+\tan \theta}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos \theta}{\sin \theta}+\dfrac{\sin \theta}{\cos \theta}}\\\\ =&\dfrac{\dfrac{1}{\cos \theta}}{\dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}}\\\\ =&\dfrac{1}{\cos \theta} \times \dfrac{\sin \theta \cos \theta}{1}\\\\ =&\sin \theta \end{aligned}$



14. Show that $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}=2 \csc x$.





$\begin{aligned} & \dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x}{1+\cos x} \times \dfrac{1-\cos x}{1-\cos x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{1-\cos ^{2} x}+\dfrac{1+\cos x}{\sin x} \\\\ =& \dfrac{\sin x(1-\cos x)}{\sin ^{2} x}+\dfrac{1+\cos x}{\sin x}\\\\ =&\dfrac{1-\cos x}{\sin x}+\dfrac{1+\cos x}{\sin x} \\\\ =&\dfrac{1-\cos x+1+\cos x}{\sin x} \\\\ =&\dfrac{2}{\sin x} \\\\ =&2 \csc x \end{aligned}$



15. Show that $\dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A}=\cot A$.





$\begin{aligned} & \dfrac{(1-\sin A)(1+\sin A)}{\sin A \cos A} \\\\ =& \dfrac{1-\sin ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos ^{2} A}{\sin A \cos A} \\\\ =& \dfrac{\cos A}{\sin A}\\\\ =&\cot A \end{aligned}$



16. Show that $\cos \theta \cot \theta+\sin \theta=\csc \theta$.





$\begin{aligned} & \cos \theta \cot \theta+\sin \theta \\\\ =& \cos \theta \cdot \dfrac{\cos \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta}{\sin \theta}+\sin \theta \\\\ =& \dfrac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta} \\\\ =& \dfrac{1}{\sin \theta} \\\\ =& \csc \theta \end{aligned}$



17. Show that $2 \cos x \cot x+1=\cot x+2 \cos x$ can be written in the form $(a \cos x-b)(\cos x-\sin x)=0$, where $a$ and $b$ are constants to be found.





$\begin{aligned} &\quad 2 \cos x \cot x+1=\cot x+2 \cos x \\\\ &\quad 2 \cos x \dfrac{\cos x}{\sin x}+1=\dfrac{\cos x}{\sin x}+2 \cos x \\\\ &\quad \text { Multiplying both sides with}\ \sin x, \\\\ &\quad 2 \cos ^{2} x+\sin x=\cos x+2 \sin x \cos x \\\\ &\quad 2 \cos ^{2} x-\cos x+\sin x-2 \sin x \cos x=0 \\\\ &\quad\cos x(2 \cos x-1)+\sin x(1-2 \cos x)=0\\\\ &\quad\cos x(2 \cos x-1)-\sin x(2 \cos x-1)=0 \\\\ &\quad(2 \cos x-1)(\cos x-\sin x)=0 \\\\ &\therefore(a \cos x-b)(\cos x-\sin x)=(2 \cos x-1)(\cos x-\sin x) \\\\ &\therefore a=2\quad \text { and }\quad b=1 \end{aligned}$

Exercise (10.4) - Trigonometric Ratios of Special Angles

Trigonometric Ratios of $30^{\circ}, 45^{\circ}$ and $60^{\circ}$

$\begin{array}{|c|c|c|c|c|c|c|} \hline \theta & \sin \theta & \cos \theta & \tan \theta & \cot \theta & \sec \theta & \csc \theta \\ \hline 30^{\circ}\left(\displaystyle\frac{\pi}{6}\right) & \displaystyle\frac{1}{2} & \displaystyle\frac{\sqrt{3}}{2} & \displaystyle\frac{\sqrt{3}}{3} & \sqrt{3} & \displaystyle\frac{2 \sqrt{3}}{3} & 2 \\ \hline 45^{\circ}\left(\displaystyle\frac{\pi}{4}\right) & \displaystyle\frac{\sqrt{2}}{2} & \displaystyle\frac{\sqrt{2}}{2} & 1 & 1 & \sqrt{2} & \sqrt{2} \\ \hline 60^{\circ}\left(\displaystyle\frac{\pi}{3}\right) & \displaystyle\frac{\sqrt{3}}{2} & \displaystyle\frac{1}{2} & \sqrt{3} & \displaystyle\frac{\sqrt{3}}{3} & 2 & \displaystyle\frac{2 \sqrt{3}}{3} \\ \hline \end{array}$

Exercise (10.4)

1. Draw a right triangle and find $\angle A$.
   
   (a) $\sin A=\displaystyle\frac{1}{2}$
   
   
   
   

   $ \begin{array}{l} \quad \quad \sin A=\displaystyle\frac{1}{2}\\\\ \quad \quad \displaystyle\frac{{\text{length of opposite side}}}{{\text{ length of hypotenuse side}}}=\displaystyle\frac{1}{2}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle}\text{.}\\\\ \therefore \quad \angle A=\text{30}^{\circ} \end{array}$

   
   (b) $\cos A=\displaystyle\frac{\sqrt{3}}{2}$
   
   
   
   

   $\begin{array}{l} \quad \quad\cos A=\displaystyle\frac{{\sqrt{3}}}{2}\\\\ \quad\quad \displaystyle\frac{{\text{length of adjacent side}}}{{\text{length of}\ \text{hypotenuse side}}}=\displaystyle\frac{{\sqrt{3}}}{2}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle}\text{.}\\\\ \therefore \quad \angle A=30^{\circ} \end{array}$

   
   (c) $\tan A=\sqrt{3}$
   
   
   
   

   $\begin{array}{l} \quad \quad\tan A=\sqrt{3}\\\\ \quad\quad \displaystyle\frac{\text{length of opposite side}}{\text{length of adjacent side}}=\sqrt{3}\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=60^{\circ} \end{array}$

   
   (d) $\cot A=1$
   
   
   
   

   $\begin{array}{l} \quad \quad\cot A=1\\\\ \quad\quad \displaystyle\frac{\text{length of adjacent side}}{\text{length of opposite side}}=1\\\\ \therefore \quad \text{the triangle is a}\ 45^{\circ} -45^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=45^{\circ} \end{array}$

   
   (e) $\sec A=\sqrt{2}$
   
   
   
   

   $\begin{array}{l} \quad \quad\sec A=\sqrt{2}\\\\ \quad\quad \displaystyle\frac{\text{length of hypotenuse side}}{\text{length of adjacent side}}=\sqrt{2}\\\\ \therefore \quad \text{the triangle is a}\ 45^{\circ} -45^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=45^{\circ} \end{array}$

   
   (f) $\csc A=2$





$\begin{array}{l} \quad \quad\csc A=2\\\\ \quad\quad \displaystyle\frac{\text{length of hypotenuse side}}{\text{length of opposite side}}=2\\\\ \therefore \quad \text{the triangle is a}\ 30^{\circ} -60^{\circ} \ \text{right triangle.}\\\\ \therefore \quad \angle A=30^{\circ} \end{array}$


For each of the right triangles $A B C$, find the indicated sides.



2. $\angle A=30^{\circ}, \quad c=30$, find $a$.





$\begin{array}{l} \quad\quad \displaystyle \frac{a}{c}=\sin A\\\\ \quad\quad \displaystyle \frac{a}{{30}}=\sin 30^{\circ} \\\\\ \quad\quad \displaystyle \frac{a}{{30}}=\displaystyle \frac{1}{2}\\\\ \quad\quad a=\displaystyle \frac{1}{2}\times 30=15 \end{array}$



3. $\angle A=60^{\circ}, \quad a=15$, find $b$.





$\begin{array}{l} \quad\quad \displaystyle \frac{b}{a}=\cot A\\\\ \quad\quad \displaystyle \frac{b}{15}=\cot 60^{\circ} \\\\\ \quad\quad \displaystyle \frac{b}{15}=\displaystyle \frac{1}{\sqrt{3}}\\\\ \quad\quad b=\displaystyle \frac{15}{\sqrt{3}}=5\sqrt{3} \end{array}$



4. $\angle B=45^{\circ}, \quad a=16$, find $c$.





$\begin{array}{l} \quad\quad \displaystyle \frac{c}{a}=\csc A\\\\ \quad\quad \displaystyle \frac{c}{16}=\csc 45^{\circ} \\\\\ \quad\quad \displaystyle \frac{c}{16}=\sqrt{2}\\\\ \quad\quad b=16\sqrt{2} \end{array}$



5. $\angle B=30^{\circ}, \quad b=8$, find $c$.





$\begin{array}{l} \quad\quad \displaystyle \frac{c}{b}=\csc B\\\\ \quad\quad \displaystyle \frac{c}{8}=\csc 30^{\circ} \\\\ \quad\quad \displaystyle \frac{c}{8}=2\\\\ \quad\quad c=16 \end{array}$



6. A ladder is placed along a wall such that it upper end is touching the top of the wall. The foot of the ladder is $5\ \text{ft}$ away from the wall and the ladder is making an angle of $60^{\circ}$ with the level of the ground. Find the height of the wall.





$\begin{array}{l}\ \ AC=5\ \text{ft}\\\\\ \ \angle A=60{}^\circ ,\ \angle C=90{}^\circ \\\\\ \ BC=?\\\\\ \ \ \text{In right }\triangle ABC,\\\\\ \ \displaystyle\frac{{BC}}{{AC}}=\tan A\\\\\ \ \displaystyle\frac{{BC}}{5}=\tan 60{}^\circ \\\\\ \ \displaystyle\frac{{BC}}{5}=\sqrt{3}\\\\\ \ BC=5\sqrt{3}\ \text{ft}\end{array}$


Find the numerical value of:


7. $\cot ^{3} 45^{\circ}+4 \sin ^{3} 30^{\circ}$.





$\begin{array}{l}\ \ \ {{\cot }^{3}}45{}^\circ +4{{\sin }^{3}}30{}^\circ \\={{(1)}^{3}}+4{{\left( {\displaystyle\frac{1}{2}} \right)}^{3}}\\={{(1)}^{3}}+4\left( {\displaystyle\frac{1}{8}} \right)\\=1+\displaystyle\frac{1}{2}\\=\displaystyle\frac{3}{2}\end{array}$



8. $\tan 60^{\circ} \cot 30^{\circ}+4 \sec ^{2} 30^{\circ}$





$\begin{array}{l}\ \ \ \tan {60}^{\circ }\cot {30}^{\circ }+4\sec^{2}30^{\circ }\\\\=\sqrt{3}\left( {\sqrt{3}} \right)+4{{\left( 2 \right)}^{2}}\\\\=3+16\\\\=19\end{array}$



9. $\tan ^{2} 45^{\circ}+\sin 30^{\circ}-\cos ^{2} 30^{\circ}+2 \tan ^{2} 60^{\circ}$.





$\begin{array}{l} \ \ \ \tan^{2} 45^{\circ} +\sin 30^{\circ} -\cos^2 30^{\circ} +2\tan^2 {60}^{\circ} \\\\ ={1}^{2}+\displaystyle\frac{1}{2}-\left(\displaystyle\frac{\sqrt{3}}{2} \right)^{2}+2\left(\sqrt{3}\right)\\\\ =1+\displaystyle\frac{1}{2}-\displaystyle\frac{3}{4}+2\sqrt{3}\\\\ =\displaystyle\frac{3+8\sqrt{3}}{4} \end{array}$



10. $\displaystyle\frac{1}{2} \sec ^{2} 30^{\circ}+\csc ^{2} 45^{\circ}-2 \tan ^{2} 30^{\circ}$.





$\begin{array}{l} \ \ \ \displaystyle \frac{1}{2}\sec^{2} 30^{\circ} +\csc^{2} 45^{\circ} -2\tan^{2} 30^{\circ} \\\\ =\displaystyle \frac{1}{2}{{\left( \displaystyle \frac{2\sqrt{3}}{3} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}-2\left( \displaystyle \frac{\sqrt{3}}{3} \right)^{2}\\\\ =\displaystyle \frac{1}{2}\left( \displaystyle \frac{4}{3} \right)+\left( 2 \right)-2\left( \displaystyle \frac{1}{3} \right)\\\\=\displaystyle \frac{2}{3}+2-\displaystyle \frac{2}{3}\\\\ =2 \end{array}$

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 $A B D C .$
   





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