It is found that on walking x metres towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60° . The height of the chimney is

Solution(By Examveda Team)

In the figure, AB is chimney and CB and DB are its shadow

$$\eqalign{ & \tan {60^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{{BC}} \cr & \Rightarrow \sqrt 3 = \frac{h}{{BC}} \cr & \Rightarrow BC = \frac{h}{{\sqrt 3 }}\,.......\,\left( {\text{i}} \right) \cr & {\text{and}} \cr & \tan {30^ \circ } = \frac{h}{{DB}} = \frac{h}{{DB + BC}} \cr & \frac{1}{{\sqrt 3 }} = \frac{h}{{x + BC}} \cr & x + BC = h\sqrt 3 \cr & \Rightarrow BC = h\sqrt 3 - x\,.......\,\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right) \cr & \frac{h}{{\sqrt 3 }} = h\sqrt 3 - x \cr & \Rightarrow \frac{h}{{\sqrt 3 }} - h\sqrt 3 = - x \cr & x = h\sqrt 3 - \frac{h}{{\sqrt 3 }} \cr & x = h\left( {\sqrt 3 - \frac{1}{{\sqrt 3 }}} \right) \cr & x = h\frac{{3 - 1}}{{\sqrt 3 }} \cr & x = \frac{{2h}}{{\sqrt 3 }} \cr & \therefore h = \frac{{\sqrt 3 }}{2}x \cr} $$