Question

A stationary balloon is observed from $3$ points $A,B$ and $C$ on the plane ground and is found that its angle of elevation from each point is $\alpha$. If $\angle ABC=\beta$ and $AC=b$, the height of the balloon is

• A. $\frac{b}{2}\tan \alpha \cdot \text{cosec}\beta$
• B. $\frac{b}{2}\tan \alpha \cdot \sin \beta$
• C. $\frac{b}{2}\cot \alpha \cdot \text{cosec}\beta$
• D. $\frac{b}{2}\cot \alpha \cdot \sin \beta$

Answer 1

The correct option is A $\frac{b}{2}\tan \alpha \cdot \text{cosec}\beta$

from sine rule,

$\frac{b}{\sin \beta }=2R\Rightarrow R=\frac{b}{2\sin \beta }..............\left(1\right)$

since, angle of elevation is same from all points , it means that distance between centre of triangle and the points is $=R$

Now, as we can see in the second(on The Right Hand side) triangle

$\tan \alpha =\frac{h}{R}\Rightarrow h=Rtan\alpha .............\left(2\right)$

Putting value of $R$ from $\left(1\right)$ in $\left(2\right)$ we get,

$h=\frac{b\tan \alpha }{2\sin \beta }=\frac{b}{2}\tan \alpha .cosec\beta$