The length of the shadow of a tower standing on level ground is found to 2x meter longer when the sun’s elevation is 30° than when it was 45 °. The height of the tower in meters is

Solution(By Examveda Team)

AB is a tower
BD and BC are its shadows and CD = 2x

$$\eqalign{ & \tan {45^ \circ } = \frac{{AB}}{{DB}} \cr & \Rightarrow 1 = \frac{h}{y} \Rightarrow y = h \cr & {\text{and}}\tan {30^ \circ } = \frac{{AB}}{{CB}} \cr & \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{2x + y}} \cr & \Rightarrow 2x + y = \sqrt 3 \,h \cr & \Rightarrow \sqrt 3 \,h - h = 2x \cr & \Rightarrow h\left( {\sqrt 3 - 1} \right) = 2x \cr & \Rightarrow h = \frac{{2x}}{{\sqrt 3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x\left( {\sqrt 3 + 1} \right)}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = x\left( {\sqrt 3 + 1} \right) \cr} $$