The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of electric pole ?

Answer: Option C

Let AB be the tower and CD be the electric pole.
Then, $$\angle ACB=60^{\circ },$$     $$\angle EDB=30^{\circ }$$     and AB = 15 m
Let CD = h
Then, BE = (AB - AE) = (AB - CD) = (15 - h)
$$\begin{array}{rl}& \frac{AB}{AC}=\tan 60^{\circ }=\sqrt{3}\\ & \Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{15}{\sqrt{3}}\\ & \text{And, }\frac{BE}{DE}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}\\ & \Rightarrow DE=\left(BE\times \sqrt{3}\right)\\ & =\sqrt{3}\left(15-h\right)\\ & \text{So, }AC=DE\\ & \Rightarrow \frac{15}{\sqrt{3}}=\sqrt{3}\left(15-h\right)\\ & \Rightarrow 3h=\left(45-15\right)\\ & \Rightarrow h=10\text{ m}\end{array}$$