From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30^°. From the bottom of the same building, the…

Let CD be the building and AB be the tower. \[\begin{array}{l} \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{\mathrm{DE}}=\frac{\mathrm{h}}{x} \\ \Rightarrow x=\mathrm{h} \sqrt{3} \\ \tan 60^{\circ}=\sqrt{3}=\frac{\mathrm{h}+15}{x} \\ \Rightarrow \mathrm{~h}+15=x \sqrt{3} \end{array} \]Solving (i) and (ii) to get $\displaystyle x=7.5 \sqrt{3} \mathrm{~m}$ or $\displaystyle \frac{15 \sqrt{3}}{2} \mathrm{~m}$ \[\text { and } \mathrm{h}=\frac{15}{2}=7.5 \mathrm{~m} \]Hence height of the tower is $\displaystyle 7.5+15=22.5 \mathrm{~m}$