Question

A round balloon of radius $a$ subtends an angle $\alpha$ at the eye of the observer while the angle of elevation of its centre is $\beta$, then height of the center of the balloon is $a\sin \beta \csc \alpha /2$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $P$ be the eye of the observer.

$PA$ and $PB$ are tangents to the round balloon.

$PX$ is the horizontal line and $CQ\perp PQ$.

Given, $\angle APB=\alpha$

Therefore, $\angle CPA=\angle CPB=\frac{\alpha }{2}$

And, $\angle CPQ=\beta$

Radius, $CA=CB=a$

Let height of the centre $C$ be $h$

In right triangle $CBP$, we have

$\sin \frac{\alpha }{2}=\frac{BC}{CP}$

$\Rightarrow \sin \frac{\alpha }{2}=\frac{a}{CP}$

$\Rightarrow CP=\frac{a}{\sin \frac{\alpha }{2}}$

$CP=a\csc \left(\frac{\alpha }{2}\right).....\left(1\right)$

In right triangle $CPQ$, we have

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

$\Rightarrow CQ=CP\sin \beta$

$\Rightarrow CQ=a\csc \left(\frac{\alpha }{2}\right)\sin \beta$ [from $\left(1\right)$]

Hence, the height is $a\sin \beta \csc \left(\frac{\alpha }{2}\right)$.

Hence, the given statement is $True$.