question_answer

From the bottom of a pole of height A, the angle of elevation of the top of a tower is \[\alpha \]. The pole subtends an angle \[\beta \]  at the top of the tower. The height of the tower is

A) \[\frac{h\sin \alpha (\alpha -\beta )}{\sin \beta }\]         

B) \[\frac{h\sin \alpha \cos (\alpha +\beta )}{\cos \beta }\]

C)  \[\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }\]              

D) \[\frac{h\sin \alpha \sin (\alpha +\beta )}{\cos \beta }\]

Correct Answer: C

Solution :

Let PQ be the tower and OA be the pole. In \[\Delta OPQ\],we have \[\tan \alpha =\frac{PQ}{OP}=\frac{PQ}{x}\] \[\Rightarrow \]                               \[PQ=x\tan \alpha \] \[\Rightarrow \]                               \[h+QR=x\tan \alpha \] \[\Rightarrow \]               \[QR=x\tan \alpha -h\]                  .. (i) In \[\Delta APQ,\]we have \[\tan (\alpha -\beta )=\frac{QR}{x}\] \[\Rightarrow \]\[\tan (\alpha -\beta )=\frac{x\tan \alpha -h}{x}\] [using Eq. (i)] \[\Rightarrow \]               \[\tan (\alpha -\beta )=\tan \alpha -\frac{h}{x}\] \[\frac{h}{x}=\tan \alpha -\tan (\alpha -\beta )\]                 \[\Rightarrow \]               \[x=\frac{h}{\tan \alpha -\tan (\alpha -\beta )}\]                                 \[\therefore \]  \[PQ=x\tan \alpha \]                 \[=\frac{h\tan \alpha }{\tan \alpha -\tan (\alpha -\beta )}=\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }\]