question_answer

AB is a straight road leading to C, the foot of a tower. A is at a distance 200 metres from C and B at 75 metres from C. If the angle of elevation of the tower at B be double the angle of elevation at A, then the height of the tower is

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            \[h=2\times 125\times \frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{5}}=100\]