• question_answer An observer finds the angle of elevation of the top of the tower from a certain point on the ground as $30{}^\circ$. If the observer moves $20\text{ }m$ towards the base of the tower, the angle of elevation of the top increases by $15{}^\circ$, find the height of the tower.

 Let AB be tower of height h From $\Delta \,ABC,\frac{AB}{BC}=\tan \,\,45{}^\circ$ $h=BC$ From $\Delta \,\,ABD,\frac{AB}{BD}=\tan \,\,30{}^\circ$ $h=\frac{BD}{\sqrt{3}}$ or $BD=\sqrt{3}\,h$ $CD=BD-BC$ $\Rightarrow$               $=\sqrt{3}\,h-h=\left( \sqrt{3}-1 \right)h$ $\Rightarrow$               $20\,m=\left( \sqrt{3}-1 \right)h$ $\Rightarrow$               $h=\frac{20}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{20\left( \sqrt{3}+1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{(1)}^{2}}}$ $=\frac{20\left( \sqrt{3}+1 \right)}{2}=10\left( \sqrt{3}+1 \right)m$