Mathematics - Heights and Distances Question with Solution | TestHub

Accoding to condition,Let, height of tower $=h$ In $△ABD,\tan 45^{\circ }=\frac{AB}{BD}\Rightarrow BD=50\mathrm{m}$ Now, In $△AC{C}^{\prime }$ $\begin{array}{rlrl}& & \tan 30^{\circ }& =\frac{A{C}^{\prime }}{C^{\prime }C}\\ \Rightarrow & & \frac{1}{\sqrt{3}}& =\frac{50-h}{50}\\ \Rightarrow & & 50& =50\sqrt{3}-h\sqrt{3}\\ \Rightarrow & & h\sqrt{3}& =50(\sqrt{3}-1)\\ & & h& =50\left(1-\frac{1}{\sqrt{3}}\right)\\ & & & =50\left(1-\frac{\sqrt{3}}{3}\right)\mathrm{m}\end{array}$