Question

A round balloon of radius r subtends an angle 2$\alpha$ at the eye of the observer while the angle of elevation of its centre is $\beta$ . Prove that the height of the contra of the balloon vertically above the horizontal level of eye is $\frac{rsin\beta }{sin\alpha }$

Answer 1

Let the height of center of the balloon above the ground be ${}^{\prime }{h}^{\prime }m$

Balloon obtends $2\alpha$ angle at the observes eye.

$\Rightarrow \angle EAD=2\alpha$

In $△ACE\&△ACD$

$\Rightarrow AE=AD$ [ length of tangents drawn from an external part to the circle are equal ]

$\Rightarrow AC=AC$ [common]

$\Rightarrow CE=CD$ [ radius of balloon ]

$\therefore △ACE\cong △ACD$ [By SSS congruence ]

$\Rightarrow \angle EAC=\angle DAC$ [By CPCT]

$\Rightarrow \angle EAC=\angle DAC=\alpha$

In right angled $△ACD,$

$\Rightarrow \sin \alpha =\frac{CD}{AC}=\frac{r}{AC}$

$\Rightarrow AC=\frac{r}{\sin \alpha }⟶\left(1\right)$

In right angled $△ACB,$

$\Rightarrow \sin \beta =\frac{BC}{AC}=\frac{h}{AC}$

$\Rightarrow AC=\frac{h}{\sin \beta }⟶\left(2\right)$

From $\left(1\right)\&\left(2\right),$ we get

$\Rightarrow h=\frac{r\sin \beta }{\sin \alpha }$

Height of the center of the balloon from the horizontal level of eye is $\frac{r\sin \beta }{\sin \alpha }.$
Hence, proved.