Question

A round balloon of radius r subtends an angle $\alpha$ at the eye of the observer while the angle of elevation of its center is $\beta$. Then the height of the balloon is

• A. $r\cos \frac{\alpha }{2}\sin \beta$
• B. $\frac{r\cos \alpha }{2\sin \beta }$
• C. $\frac{r\cos \alpha }{2\cos \beta }$
• D. $r\sin \alpha \sin \beta$

Answer 1

Answer: (A)

$r\cos \frac{\alpha }{2}\sin \beta$

Let $O$ be the center of the balloon, $P$ be the eye of the observer, and let $\angle APB=\alpha$

OL will be the height of the ballon if we draw a $\perp$ from point $O$.

$\angle APO=\angle BPO=\frac{\alpha }{2}$ $(\because △AOP\cong △BOP)$

We know, $sin\left(\Theta \right)=Opposite/Hypotenuse$

In $△AOP$

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

or $OP=r\cos \frac{\alpha }{2}\to \left(1\right)$

In $△OPL$

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

$\Rightarrow OL=OP\sin \beta$

$\Rightarrow OL=r\cos \frac{\alpha }{2}\sin \beta$ (from(1))

$\therefore A$ is the answer.