From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?

Answer: Option C

Consider the diagram shown above. AC represents the hill and DE represents the pole
Given that AC = 100 m
∠XAD = ∠ADB = 30° (∵ AX || BD )
∠XAE = ∠AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h,
AB = (100 - h)   (∵ AC = 100 and BC = h),
BD = CE
$$\begin{array}{rl}& \tan 60^{\circ }=\frac{AC}{CE}\\ & \Rightarrow \sqrt{3}=\frac{100}{CE}\\ & \Rightarrow CE=\frac{100}{\sqrt{3}}......\left(1\right)\\ & \tan 30^{\circ }=\frac{AB}{BD}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{100-h}{BD}\end{array}$$
$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100-h}{\left(\frac{100}{\sqrt{3}}\right)}$$      (∵ BD = CE and substituted the value of CE from eq. 1)
$$\begin{array}{rl}& \Rightarrow \left(100-h\right)=\frac{1}{\sqrt{3}}\times \frac{100}{\sqrt{3}}\\ & =\frac{100}{3}\\ & =33.33\\ & \Rightarrow h=100-33.33=66.67\text{m}\end{array}$$
i.e., the height of the pole = 66.67 m