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 $r\sin \beta \mathrm{cosec}\frac{\alpha }{2}$.

Answer 1

Let $O$ be the centre of the ballon and $P$ be the eye of observer and $\angle APB=\alpha$ be the angle subtended by the balloon at the eye of observer.

$OA=OB=r=radiusofballoon$

In $△OAP,$

$\sin \frac{\alpha }{2}=\frac{OA}{OP}$

$OP=r\text{cosec}\frac{\alpha }{2}$ ....(1)

Now,

In $△OPL,$

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

$OP\sin \beta =OL$

So,

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

Hence,

The height of the centre of the balloon is $OL=r\text{cosec}\frac{\alpha }{2}\sin \beta$.