Question

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

Answer 1

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Let the height of the center of the balloon the ground be h m

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

$\therefore \angle EAD=\theta$

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

$AE=AD$ (Length of tangent drawn from an external point to the circle are equal)

$AC=AC$ (common)

$CE=CD$ (Radius of the circle)

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

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

$\therefore \angle EAC=\angle DAC=\frac{\theta }{2}$

In right $\Delta ACD$

$sin\frac{\theta }{2}=\frac{CD}{AC}$

$\Rightarrow sin\frac{\theta }{2}=\frac{r}{AC}$

$\Rightarrow AC=\frac{r}{sin\frac{\theta }{2}}$

$\Rightarrow AC=rcosec\frac{\theta }{2}-------\left(1\right)$

In right $\Delta ACB$

$sin\varphi =\frac{BC}{AC}$

$\therefore sin\varphi =\frac{h}{cosec\frac{\theta }{2}}$

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