From an aeroplane above a straight road the angle of depression of two positions at a distance $$20\,m$$  apart on the road are observed to be $${30^ \circ }$$ and $${45^ \circ }.$$ The height of the aeroplane above the ground is :

Solution :

$$\eqalign{ & {\text{In }}\,\Delta \,ABC,\tan {45^ \circ } = \frac{{AB}}{{AC}} = \frac{h}{x};1 = \frac{h}{x} \cr & h = x\,\,\,.....\left( {\text{i}} \right) \cr & {\text{In }}\Delta \,ABD, \cr & \tan {30^ \circ } = \frac{{AB}}{{BD}};\,\,\,\frac{1}{{\sqrt 3 }} = \frac{h}{{x + 20}} \cr & x + 20 = \sqrt 3 h;\,\,\,h + 20 = \sqrt 3 h \cr & 20 = \left( {\sqrt 3 - 1} \right)h;\,\,\,h = \frac{{20}}{{\sqrt 3 - 1}} \cr & = \frac{{20}}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} \cr & = \frac{{20\left( {\sqrt 3 + 1} \right)}}{2} = 10\left( {\sqrt 3 + 1} \right)m \cr} $$
Hence, the height is $$10\left( {\sqrt 3 + 1} \right)m$$