The angle of elevation of an aeroplane as observed from a point 30 m above the transport water-surface of lake is 30° and the angle of depression of the image of the aeroplane in the water of the lake is 60°. The height of the aeroplane from the water-surface of the lake is

Solution (By Examveda Team)

$$\eqalign{ & \Delta ABE \cr & \tan {30^ \circ } = \frac{{AB}}{{EB}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{x} \cr & \Rightarrow x = \sqrt 3 \times h \cr & \Delta EBD \cr & \tan {60^ \circ } = \frac{{BD}}{{EB}} \cr & \sqrt 3 = \frac{{h + 30 + 30}}{x} \cr & \sqrt 3 = \frac{{h + 60}}{{\sqrt 3 h}} \cr & 3h = h + 60 \cr & 2h = 60 \cr & h = 30{\text{m}} \cr & {\text{Height from water surface}} \cr & = 30 + 30 \cr & = 60{\text{m}} \cr} $$