A) \[\frac{2\sin \alpha \sin \beta }{\sin \left( \beta -\alpha \right)}\]
B) \[\frac{\sin \alpha \sin \beta }{\cos \left( \beta -\alpha \right)}\]
C) \[\frac{2\sin \left( \beta -\alpha \right)}{\sin \alpha \sin \beta }\]
D) \[\frac{\cos \left( \beta -\alpha \right)}{\sin \alpha \sin \beta }\]
Correct Answer: A
Solution :
Let AB be the tower of height 'h'. Given : In\[\ln \Delta ABP\] \[\tan \alpha =\frac{AB}{PB}\]or\[\frac{\sin \alpha }{\cos \alpha }=\frac{h}{x+2}\] \[\Rightarrow \]\[(x+2)sin\alpha =h\,cos\alpha \] \[\Rightarrow \]\[h=\frac{x\sin \alpha +2\sin \alpha }{\cos \alpha }\] ?(1) Now, \[\ln \Delta ABC,\tan \beta =\frac{AB}{BC}\] \[\Rightarrow \]\[\frac{\sin \beta }{\cos \beta }=\frac{h}{x}\Rightarrow x=\frac{h\cos \beta }{\sin \beta }\] ?(2) Putting the value of x in eq. (2) to eq. (1), we get\[h=\frac{\frac{h\cos \beta \sin \alpha }{\sin \beta }+\frac{2\sin \alpha }{1}}{\cos \alpha }\] \[\Rightarrow \]\[h=\frac{h\cos \beta .\sin \alpha +2\sin \alpha \sin \beta }{sin\beta .cos\alpha }\] \[\Rightarrow \]\[h(sin\beta .cos\alpha -cos\beta .sin\alpha )=2sin\alpha .sin\beta \] \[\Rightarrow \]\[h[sin(\beta -\alpha )]=2sin\alpha .sin\beta \] \[\Rightarrow \]\[h=\frac{2\sin \alpha .\sin \beta }{\sin (\beta -\alpha )}\]You need to login to perform this action.
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