# The angle of elevation of the top of a tower from a certain point is 30°. If the observed moves 20 m towards the tower, the angle of elevation the angle of elevation of top of the tower increases by 15°. The height of the tower is

A. 17.3 m B. 21.9 m C. 27.3 m D. 30 m Answer: Option C
Let AB be the tower and C and D be the point of observation Given, $\mathrm{CD}=20\mathrm{~m}\text{ And }\angle\mathrm{BCA}=30^{\circ}$ $\text { and } \angle \mathrm{BDA}=30+15=45^{\circ}$ Let Height of tower is h In Triangle BAD $\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{AD}} \Rightarrow 1=\frac{\mathrm{h}}{\mathrm{AD}} \Rightarrow \mathrm{AD}=\mathrm{h}$ In triangle BAC $\begin{array}{l} \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}}(\mathrm{AC}=\mathrm{CD}+\mathrm{AD}) \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{20+\mathrm{h}} \\ \Rightarrow \sqrt{3} \mathrm{~h}=20+\mathrm{h} \Rightarrow \sqrt{3} \mathrm{~h}-\mathrm{h}=20 \\ \Rightarrow \mathrm{h}(1.732-1)=20 \\ \Rightarrow \mathrm{h}=\frac{20}{0.732}=27.3~m\end{array}$