question_answer

A round balloon of radius\[r\]subtends an angle \[\alpha \]at the eye of the observer, while the angle of elevation of its centre is\[\beta \]. The height of the centre of balloon is

A) \[r\cos ec\alpha \sin \frac{\beta }{2}\]  

B) \[r\sin \alpha \cos ec\frac{\beta }{2}\]

C) \[r\sin \frac{\alpha }{2}\cos ec\beta \]

D)        \[r\cos ec\frac{\alpha }{2}\sin \beta \]

Correct Answer: D

Solution :

In\[\Delta \,\,APC\],\[\sin (\angle PAC)=\frac{CP}{AC}\] \[\Rightarrow \]\[AC=\frac{r}{\sin \frac{\alpha }{2}}=r\,\,\cos ec\frac{\alpha }{2}\]                          ... (i) Again in\[\Delta ABC\],                 \[\sin \beta =\frac{BC}{AC}\] \[\Rightarrow \]               \[BC=AC\,\,\sin \beta \] \[\Rightarrow \]               \[H=r\sin \beta \cos ec\left( \frac{\alpha }{2} \right)\]                                                 [from Eq, (i)]