Question

A round balloon of radius ‘r’ subtends an angle $\theta$ at the eye of the observer, while angle of elevation of its center is $\varphi$. Find the height of the center of the balloon.

Answer 1

Let center of balloon of radius $r$ of $O$ and $A$ is the position of observer. let $AP$ and $AQ$ are too tangents to the balloon.
$OB=h$ m
Balloon of radius $r$, subtends an angle $\theta$ and angle of elevation of its center is $\varphi$.
Then
$\angle PAQ=\theta$
and $\angle PAO=\angle QAO=\theta /2$
$OP=r$ and $\angle PAB=\varphi$
$\angle APO=\angle AQO={90}^0$
[$because$ Radius and tangents are perpendicular to each other]
From right angled $\Delta OAB$,
$\sin \varphi =\frac{OB}{AO}$
$\sin \varphi =\frac{h}{AO}$
$h=AO\sin \varphi$ ….(i)
From right angled $\Delta AOP$,
$\sin \frac{\theta }{2}=\frac{OP}{AO}$
$\sin \frac{\theta }{2}=\frac{r}{AO}$
$AO=r.cosec\frac{\theta }{2}$ ……(ii)
Put the value of $AO$ in equation (i),
$h=r.cosec\frac{\theta }{2}\sin \varphi$
Hence, height of the center of the balloon (h) = $r.cosec\frac{\theta }{2}\sin \varphi$