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)] \[\Rightarrow \] \[\tan (\alpha -\beta )=\tan \alpha -\frac{h}{x}\] \[\frac{h}{x}=\tan \alpha -\tan (\alpha -\beta )\] \[\Rightarrow \] \[x=\frac{h}{\tan \alpha -\tan (\alpha -\beta )}\] \[\therefore \] \[PQ=x\tan \alpha \] \[=\frac{h\tan \alpha }{\tan \alpha -\tan (\alpha -\beta )}=\frac{h\sin \alpha \cos (\alpha -\beta )}{\sin \beta }\]You need to login to perform this action.
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