The angle of elevation of the top of a tower from a certain point is 30°. If the observer moves 40 m towards the tower, the angle of elevation of the top of the tower increases by 15°. The height of the tower is:

Answer: Option D

Let DC be the tower and A and B be the positions of the observer such that AB = 40 m
We have ∠DAC = 30°, ∠DBC = 45°
Let DC = h
$$\begin{array}{rl}& \tan 30^{\circ }=\frac{DC}{AC}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{AC}\\ & \Rightarrow AC=h\sqrt{3}.........\left(\text{i}\right)\\ & \tan 45^{\circ }=\frac{DC}{BC}\\ & \Rightarrow 1=\frac{h}{BC}\\ & \Rightarrow BC=h........\left(\text{ii}\right)\\ & \text{We know that,}\\ & AB=\left(AC-BC\right)\\ & \Rightarrow 40=\left(AC-BC\right)\end{array}$$
$$\Rightarrow 40=\left(h\sqrt{3}-h\right)$$   [∵ from (1) and (ii)]
$$\begin{array}{rl}& \Rightarrow 40=h\left(\sqrt{3}-1\right)\\ & \Rightarrow h=\frac{40}{\left(\sqrt{3}-1\right)}\\ & =\frac{40}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)}\\ & =\frac{40\left(\sqrt{3}+1\right)}{\left(3-1\right)}\\ & =\frac{40\left(\sqrt{3}+1\right)}{2}\\ & =20\left(\sqrt{3}+1\right)\\ & =20\left(1.73+1\right)\\ & =20\times 2.73\\ & =54.6m\end{array}$$