Question

A spherical balloon of radius $r$ subtends an angle $\alpha$ at the eye of an observer, while the angle of elevation of its centre is $\beta$. The height of the centre of the balloon is:

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

Answer 1

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

Let O be the center of the balloon of radius r and P be the eye of an observer

Let PA and PB be the tangents from P to the balloon.

Then, $\angle APX=\alpha$

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

Let OL be the tar drawn from center O on PX

$\angle OPL=\beta$

In $△OAP$

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

$OP=r\csc \frac{\alpha }{2}$

In $△OPL,\sin \beta =\frac{OL}{OP}$

$OL=OP\sin \beta =r\csc \frac{\alpha }{2}.\sin \beta$

$h=r\csc \frac{\alpha }{2}\sin \beta$