question_answer

A balloon, whose radius is r, subtends an angle a at the eye of an observer, when the angle of elevation of its centre is P. The height of its centre is

A) \[r\,\text{cosec}\frac{\alpha }{2}\sin \beta \]

B) \[\frac{r\,\text{cosec}\,\alpha }{2\sin \beta }\]

C) \[\frac{r\,\text{cosec}\,\alpha }{2\cos \beta }\]

D) \[r\sin \alpha \sin \beta \]

Correct Answer: A

Solution :

 If P is the eye, then                 \[\angle CPQ=\beta ,\,\angle CPA=\angle CPB=\frac{\alpha }{2}\] Let CR = h be the height of centre of balloon From right angled \[\Delta \,CAP,\]                 \[\sin \frac{\alpha }{2}=\frac{CA}{CP}=\frac{r}{CP}\] \[\therefore \]  \[CP=r\,\text{cosec}\,\frac{\alpha }{2}\] Also, from right angled \[\Delta \,CPR,\]                 \[\sin \beta =\frac{h}{CP}=\frac{h}{r\,\text{cosec}\frac{\alpha }{2}}\] \[\therefore \]  \[h=r\,\,\text{cosec}\frac{\alpha }{2}\sin \beta \]