From the top of a hill 200 m high the angle of depression of the top and the bottom of a tower are observed to be 30° and 60°. The height of the tower is (in m);

Solution (By Examveda Team)

$$\eqalign{ & AB = {\text{hill}} = 200{\text{ metre}} \cr & CD = {\text{tower}} \cr & {\text{In }}\Delta \,APC \cr & \tan {30^ \circ } = \frac{{AP}}{{PC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{AP}}{{PC}} \cr & \Rightarrow AP:PC = 1:\sqrt 3 \,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{In }}\Delta \,ABD \cr & \tan {60^ \circ } = \frac{{AB}}{{BD}} \cr & \sqrt 3 = \frac{{AB}}{{BD}} \cr & \Rightarrow AB:BD = \sqrt 3 :1\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & PB = CD{\text{ and }}PC = BD \cr & {\text{Now}} \cr} $$
\[\begin{array}{*{20}{c}} {AB}&:&{BD}&:&{AP} \\ {\sqrt 3 }&:&1&{}&{} \\ {}&{}&{\sqrt 3 }&:&1 \\ 3&:&{\sqrt 3 }&:&1 \end{array}\]
$$\eqalign{ & CD = PB \Rightarrow AB - AP \cr & CD = 3 - 1 \cr & CD = 2{\text{ units}} \cr & AB = 3{\text{ units}} = 200{\text{ metre}} \cr & CD = 2{\text{ units}} = \frac{{200}}{3} \times 2 = 133\frac{1}{3}\,{\text{metre}} \cr} $$