$A$ spherical balloon of radius $r$ subtends an angle $\theta$ at the eye of an observer. If the angle of elevation of its centre is $\phi$,find the height of the centre of the balloon.

(N/A) Let $O$ be the centre of the balloon,$OP = r$ be its radius,and $A$ be the position of the observer's eye. The angle subtended by the balloon at the observer's eye is $\angle PAQ = \theta$. The line $AO$ bisects $\angle PAQ$,so $\angle OAP = \frac{\theta}{2}$.
The angle of elevation of the centre $O$ is $\angle OAB = \phi$,where $B$ is the point on the ground directly below $O$. Let $h$ be the height of the centre of the balloon,so $OB = h$.
In the right-angled triangle $\triangle OAP$ (where $\angle OPA = 90^\circ$ because the line of sight is tangent to the sphere),we have:
$\sin(\angle OAP) = \frac{OP}{OA}$
$\sin\left(\frac{\theta}{2}\right) = \frac{r}{OA}$
$OA = \frac{r}{\sin(\frac{\theta}{2})} = r \operatorname{cosec}\left(\frac{\theta}{2}\right)$ .....$(1)$
In the right-angled triangle $\triangle OAB$ (where $\angle OBA = 90^\circ$),we have:
$\sin(\angle OAB) = \frac{OB}{OA}$
$\sin \phi = \frac{h}{OA}$
$h = OA \sin \phi$ .....$(2)$
Substituting the value of $OA$ from equation $(1)$ into equation $(2)$:
$h = r \operatorname{cosec}\left(\frac{\theta}{2}\right) \sin \phi$
$h = r \sin \phi \operatorname{cosec}\left(\frac{\theta}{2}\right)$