A cone is circumscribed about a sphere of radius r and  $\theta$ is the semi-vertical angle of cone. If the volume of the cone is minimum then  $9\sin \theta =$____

$h=r+\frac{r}{\sin \theta }$

Base  radius of $cone = R=(r+\frac{r}{\sin \theta })\tan \theta$  

$$\begin{gathered} \therefore VolumeV=\frac{1}{3}\pi R^{2}h \\ =\frac{\pi r^3}{3}\frac{{(1+\sin \theta )}^3}{Sin\theta \text{\hspace{0.17em}}Co{s}^2\theta } \\ \frac{dV}{d\theta }=\frac{\pi r^3}{3}\frac{{(1+\sin \theta )}^3(3Sin\theta -1)}{Si{n}^2\theta Co{s}^3\theta } \\ For Min/max, \Rightarrow \frac{dV}{d\theta }=0=1Sin\theta =1/3. \end{gathered}$$