Question

A round ballon of radius r subtends an $\alpha$ at the eye of the observer while the an elevation of its centre $\beta$. The height of centre of the balloon is

• A. $rsin\beta cosec\frac{\alpha }{2}$
• B. $rsin\beta /2cos\alpha /2$
• C. $rsin\beta /2cos\alpha$
• D. $rsin\beta cos\alpha$

Answer 1

The correct option is A

$rsin\beta cosec\frac{\alpha }{2}$

Let O be the central of the circle with radius r and P be the observer. Also PA and PB be lengts from P to the balloon such that $\angle APB=\alpha$
In $\Delta OAP.$
$sin\frac{\alpha }{2}=\frac{OA}{OP}\Rightarrow sin\frac{\alpha }{2}=\frac{r}{OP}$
$\Rightarrow OP=rcosec\frac{\alpha }{2}$ ...(i)
In $\Delta OPL$. we have $sin\beta =\frac{OL}{OP}$
$\Rightarrow OL=OPsin\beta =rcosec\frac{\alpha }{2}sin\beta$
[From Eq. (i)]
$\therefore$ Height of the centre of the balloon is $rsin\beta cosec\frac{\alpha }{2}$