A) 125 m
B) 120 m
C) 115m
D) 100 m
Correct Answer: D
Solution :
Here, \[\angle CBP=\angle A+\angle AP\] or \[2\alpha =\alpha =\alpha +\angle APB\] or \[\angle APB=\alpha \] \[\therefore \] \[AB=BP=125m\] From \[\Delta \,BCP,\] \[\cos \,2\alpha =\frac{75}{125}\] or \[\cos \,2\alpha =\frac{3}{5}\] \[\therefore \] \[\cos \,\alpha =\frac{2}{\sqrt{5}}\] and \[\sin \alpha =\frac{1}{\sqrt{5}}\] Again, \[\frac{h}{125}=\sin \,2\alpha =2\,\sin \alpha \,\cos \alpha \] or \[h=2\times 125\times \frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{5}}=100\]You need to login to perform this action.
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