Question

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

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

Answer 1

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

$\begin{array}{c}PXisthehorizontallineandCQ\perp PQ.\\ Itisgiventhat\\ \angle APB=\alpha \\ \angle CPA=\angle CPB=\frac{\alpha }{2}\\ and,\angle CPX=\beta \\ LetheightofthecentreCbehmandCA=CB=r\\ InrighttriangleCBPwehave\\ \sin \left(\frac{\alpha }{2}\right)=\frac{BC}{CP}\\ \sin \left(\frac{\alpha }{2}\right)=\frac{r}{CP}\\ CP=\frac{r}{\sin \left(\frac{\alpha }{2}\right)}\\ CP=r\cos ec\frac{\alpha }{2}\\ InrighttriangleCPQ,wehave\\ \sin \beta =\frac{CQ}{CP}\\ CQ=CP\sin \beta \\ CQ=rco\sec \frac{\alpha }{2}\sin \beta \\ Hence,thehieghtofthecentre\\ =r\sin \beta \cos ec\frac{\alpha }{2}\\ Hence,optionBisthecorrectanswer.\end{array}$