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 $\beta$. Prove that the height of the center of the balloon is r sin $\beta$ cosec $\alpha$/2.

Answer 1

REF.Image

Let P be age of observer. PA & PB are tangents to round balloon,

$\angle APB=\alpha$

$\therefore \angle CPA=\angle CPB=\frac{\alpha }{2}$

$CA=CB=r$ and CQ= height = h

In $\Delta CBP$

$sin\left(\frac{\alpha }{2}\right)=\frac{BC}{CP}=\frac{r}{CP}$

$\therefore CP=\frac{r}{sin\left(\frac{\alpha }{2}\right)}=rcosec\left(\frac{\alpha }{2}\right)$

In $\Delta CPQ$,

$sin\beta =\frac{CQ}{CP}$ $\therefore CQ=CPsin\beta$

$CQ=rcoesc\left(\frac{\alpha }{2}\right),sin\beta$ [from (1)]

Hence,

height = h= CQ = $rsin\beta ,cosec\left(\frac{\alpha }{2}\right)$