Question

From the top of a cliff, 200 m high, the angle of depression of the top and bottom of a tower are observed to be ${30}^{\circ }$ and ${60}^{\circ }$, find the height of the tower (in fraction; in meter)___

Answer 1

Let the height of the tower be h m, then
In $\Delta PBM,tan{60}^{\circ }=\frac{PM}{BM}\sqrt{3}=\frac{200}{BM}BM=\frac{200}{\sqrt{3}}m=AL\left(also\right)$
Now, in $\Delta ALP$
$\begin{array}{cc}& tan{30}^{\circ }=\frac{PL}{AL}\\ & \frac{1}{\sqrt{3}}=\frac{PL}{\left(\frac{200}{\sqrt{3}}\right)}\\ \Rightarrow & PL=\frac{200}{3}\\ \therefore & LM=PM-PL=200-\left(\frac{200}{3}\right)\\ & LM=\frac{400}{3}\\ But& LM=AB=h\end{array}$
$\therefore$ Hight of the tower $\left(h\right)=\frac{400}{3}m$