Question

The radius of the base of a right circular cone of semi-vertical angle $\alpha$ is $r$. Show that its volume is $\frac{1}{3}\pi r^3\cot \alpha$ and curved surface area is $\pi r^{2}cosec\alpha$

Answer 1

Let height of cone be 'h', then in $△OBA$

$\tan \alpha =\frac{OA}{OB}=\frac{r}{h}\Rightarrow h=\frac{r}{\tan \alpha }=r\cot \alpha$

Volume $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2.r\cot \alpha =\frac{1}{3}\pi r^3\cot \alpha$

CSA$=\pi rl=\pi r\sqrt{r^2+h^2}=\pi r=\sqrt{r^2+{\left(r\cot \alpha \right)}^2}$

$=\pi r\sqrt{r^2+r^2{\cot }^2\alpha }=\pi r\sqrt{r^2(1+{\cot }^2\alpha )}=\pi r\sqrt{r^2{\csc }^2\alpha }$

$=\pi r.r\csc \alpha =\pi r^2\csc \alpha$