question_answer

The angle of elevation of the top of a TV tower from three points A, B and C in a straight line through the foot of the tower are \[\alpha ,2\alpha \] and \[3\alpha \]respectively. If \[AB=a,\] then height of the tower is

A)  \[a\text{ }tan\text{ }\alpha \]          

B)  \[a\text{ }sin\text{ }\alpha \]

C)  \[a\text{ }sin\text{ 2}\alpha \]   

D)  \[a\text{ }sin\text{ 3}\alpha \]

Correct Answer: C

Solution :

In  \[\Delta ABE,\] \[\angle BAE=\angle AEB\]                 \[\therefore \]  \[AB=BE\]                                 In  \[\Delta \,DCE,\]                 \[\sin \,3\alpha =\frac{h}{CE}\] \[\Rightarrow \]               \[\sin \,3\alpha =\frac{h}{\alpha \,\sin \,2\alpha /\sin \,3\alpha }\] [from Eq. (i)] \[\Rightarrow \]               \[h=a\,\sin \,2\alpha \]