Question

A balloon of radius r makes an angle at the eye of an observer and the angle of elevation of its centre is $\beta$. The height of its centre from the ground level is given by

• A. $rcos\frac{\beta }{2}sec\alpha$
• B. $rcos\beta sec\frac{\alpha }{2}$
• C. $rsin\frac{\alpha }{2}cosec\beta$
• D. $rsin\beta cosec\frac{\alpha }{2}$

Answer 1

The correct option is D $rsin\beta cosec\frac{\alpha }{2}$

$\frac{r}{x}=\sin \frac{\alpha }{2}$

$x=\frac{r}{\sin \frac{\alpha }{2}}=r\csc \frac{\alpha }{2}$

$\frac{h}{x}=\sin \beta$

$h=x\sin \beta =r\sin \beta \csc \frac{\alpha }{2}$