Question

A balloon of radius $r$ subtends an angle $\alpha$ at the eye of an observer and the elevation of the centre of the balloon from the eye is $\beta$, the height $h$ of the centre of the balloon is given by

• A. $\frac{r\sin \beta }{\sin \alpha }$
• B. $r\sin \beta \sin \alpha$
• C. $\frac{r\sin \beta }{\sin \left(\alpha /2\right)}$
• D. $\frac{r\sin \beta }{\sin \left(\beta /2\right)}$

Answer 1

The correct option is C $\frac{r\sin \beta }{\sin \left(\alpha /2\right)}$

$\begin{array}{c}\angle APO=\angle BPO=\frac{\alpha }{2}\\ \Rightarrow In\Delta APO,\sin \frac{\alpha }{2}=\frac{OA}{OP},OP=\left(\gamma \cos ec\frac{\alpha }{2}\right)\\ andIn\Delta OPG,\sin \beta =\frac{OG}{OP},OG=OP,\sin \beta \\ \Rightarrow OG=\left(\gamma \cos ec\frac{\alpha }{2}.\sin \beta \right)\\ \Rightarrow Thus,heightofthecentreoftheballoonis\left(\gamma \sin \beta .\cos ec\frac{\alpha }{2}\right)\end{array}$