Question

A round balloon of radius $r$ subtends an angle $\theta$ at the eye of the observer, while the angle of elevation of its center is $\varphi$. Prove that the height of the centre of balloon is $r\sin \varphi \csc \frac{\theta }{2}$

Answer 1

Let the height of centre of the balloon be hm.

Given, balloon subtends an angle $\theta$ at the observes eye.

$\therefore \angle EAD=\theta$

In $\Delta ACE$ and $\Delta ACD$,

AE$=$AD(length of tangents drawn from a external point to the ob are equal)

AC$=$AC(common), CE$=$CD(radius of ob)

$\therefore \Delta$ACE $\cong \Delta$ACD(SSS congruence sub)

$\Rightarrow \angle EAC=\angle DAC\left(CPCT\right)$

$\therefore EAC=DAC=\theta /2$

In right angled $\Delta$ACD, $\sin \theta /2=CD/AC$

$\therefore \sin \theta /2=a/AC$

$\Rightarrow AC=\frac{a}{\sin \frac{\theta }{2}}=acosec\left(\frac{\theta }{2}\right)$ …………..(i)

In right angles $\Delta ACB$, $\sin \varphi =BC/AC$

$\therefore \sin \varphi =\frac{h}{acosec\left(\frac{\theta }{2}\right)}$ using (i)

$\Rightarrow h=a\sin \varphi cosec\left(\frac{\theta }{2}\right)$

Thus, the height of the centre of the balloon is $a\sin \varphi cosec\frac{\theta }{2}$

Hence proved.