Question

A spherical balloon of radius $r$ subtends an angle $\angle$ at the eye of an observer, while the angle of elevation of its centre is $\beta$. What is the height of the centre of the balloon (neglecting the height of the observer)?

• A. $\frac{r\sin \beta }{\sin \left(\frac{\alpha }{2}\right)}$
• B. $\frac{r\sin \beta }{\sin \left(\frac{\alpha }{4}\right)}$
• C. $\frac{r\sin \left(\frac{\beta }{2}\right)}{r\sin \alpha }$
• D. $\frac{r\sin \alpha }{\sin \left(\frac{\beta }{2}\right)}$

Answer 1

The correct option is A $\frac{r\sin \beta }{\sin \left(\frac{\alpha }{2}\right)}$

$\to$ Let ${}^{\prime }{O}^{\prime }$ be the centre of the balloon, $P$ be the eye of the observer, and $\angle APB$ be the angle subtended by the balloon at the eye of the observer.

$\therefore \angle APB=\alpha$

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

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

$\to \sin \frac{\alpha }{2}=\frac{r}{OP}$ or $OP=r\text{cosec}\frac{\alpha }{2}$

In $OPL$

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

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

$=\frac{r\sin \beta }{\sin \left(\frac{\alpha }{2}\right)}$.