From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
A. 25 m
B. 50 m
C. 75 m
D. 100 m
Answer: Option B
Solution(By Examveda Team)
Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at CLet angle be Q and CD = 25 m
$$\eqalign{ & {\text{Let}}\,AB = h \cr & CE \,\, || \,\, DB \cr & \therefore EC = DB = x{\text{ }}\left( {{\text{suppose}}} \right) \cr & EB = CD = 25 \cr & \therefore AE = h - 25 \cr & {\text{Now in right }}\Delta CDB, \cr & \tan \theta = \frac{{CD}}{{DB}} = \frac{{25}}{x}\,......\left( {\text{i}} \right) \cr & {\text{and in right }}\Delta CAE \cr & \tan \theta = \frac{{AE}}{{CE}} = \frac{{h - 25}}{x}\,......\left( {{\text{ii}}} \right) \cr & {\text{From}}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \frac{{25}}{x} = \frac{{h - 25}}{x} \cr & \Rightarrow 25 = h - 25 \cr & \Rightarrow h = 25 + 25 = 50 \cr & \therefore {\text{Height of tower}} = 50\,m \cr} $$
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