From a lighthouse the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the lighthouse is h metres, the distance between the ships is

Solution(By Examveda Team)

Let AB be lighthouse and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB = x and QB = y

$$\eqalign{ & {\text{Now in right }}\Delta APB \cr & \tan \theta = \frac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \frac{{AB}}{{PB}} \cr & \Rightarrow \tan {45^ \circ } = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \cr & \Rightarrow x = h\,............(i) \cr & {\text{Similarly in right }}\Delta AQB, \cr & \tan {30^ \circ } = \frac{{AP}}{{QB}} = \frac{h}{y} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{y} \cr & \Rightarrow y = \sqrt 3 \,h\,..............(ii) \cr & {\text{Adding (i) and (ii)}} \cr & \therefore PQ = x + y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = h + \sqrt 3 \,h \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\sqrt 3 + 1} \right)\,h \cr} $$