The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart, the height of the light house is

Solution(By Examveda Team)

Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively

∠ACB = ∠XAC = 30° , ∠ADB = ∠YAD = 45° and
CD = 100m
Let AB =h and CB = x then BC = (100 - x)m
$$\eqalign{ & {\text{Now in }}\,\Delta ACB, \cr & \tan \theta = \frac{{AB}}{{CB}} \cr & \tan {30^ \circ } = \frac{h}{x} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 h ......{\text{(i)}} \cr & {\text{Similarly in }}\,\Delta ADB \cr & \tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & \Rightarrow 1 = \frac{h}{{100 - x}} \cr & \Rightarrow x = 100 - h \,......{\text{(ii)}} \cr & {\text{From (i) and (ii)}} \cr & \sqrt 3 h = 100 - h \cr & \Rightarrow (\sqrt 3 - 1)h = 100 \cr & h = \frac{{100}}{{\sqrt 3 + 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\, = \frac{{100\left( {\sqrt 3 - 1} \right)}}{2} \cr & \,\,\,\,\,\, = 50\left( {\sqrt 3 - 1} \right) \cr} $$
∴ height of light house $$ = 50\left( {\sqrt 3 - 1} \right)m$$