Question

A round balloon of radius r subtends an angle $\alpha$ at the eye of the observer, while the angle of elevation of its centre is $\beta$. Prove that the height of the centre of the balloon is $r\sin \beta \text{csc}\frac{\alpha }{2}$.

Answer 1

Let $O$ be the centre of the balloon and $P$ be the eye of the observer

And $\angle APB$ be the angle subtended by the balloon

$\therefore \angle APB=\alpha$

$\therefore \angle APO=\angle BPO=\frac{\alpha }{2}$

In $△OAP,\sin \frac{\alpha }{2}=\frac{OA}{OP}$

$⟹\sin \frac{\alpha }{2}=\frac{r}{OP}$

$⟹OP=r\text{cosec}\frac{\alpha }{2}-\left(1\right)$

In $△OPC$

$\sin \beta =\frac{OL}{OP}$ or $OL=OP\sin \beta$

$\therefore OL=r\text{cosec}\frac{\alpha }{2}\sin \beta$