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 $\beta$. Find the height of the centre of balloon.

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

Answer 1

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


Given that $\angle APB=\alpha$
The lenghts of the tangents from an external point are equal.
$\Delta APO$ and $\Delta BPO$ are congruent.
Hence, $APO=\angle BPO=\frac{\alpha }{2}$
Consider the triangle $\Delta APO$;
$sin\frac{\alpha }{2}=\frac{OA}{OP}$
$\Rightarrow OP=\frac{OA}{sin\frac{\alpha }{2}}$
$\Rightarrow OP=rcosec\frac{\alpha }{2}\cdots \left(1\right)$
Now consider the triangle, $\Delta OPG$ :
$sin\beta =\frac{OG}{OP}$
$\Rightarrow OG=OPsin\beta$
$\Rightarrow OG=rcosec\frac{\alpha }{2}sin\beta$ [from equation (1)]
Thus, height of the centre of the balloon is $rsin\beta cosec\frac{\alpha }{2}$