The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°,the elevation changes to 60°. What is the approximate height of the mountain?

Answer: Option D

Let A be the foot and C be the summit of a mountain.
Given that ∠CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that ∠DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., ∠CDE = 60°
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta ABC,\\ & \tan 45^{\circ }=\frac{CB}{AB}\end{array}$$
$$\Rightarrow 1=\frac{x}{AB}$$    [∵ CB = x(the height of the mountain)]
$$\begin{array}{rl}& \Rightarrow AB=x......\left(1\right)\\ & \text{From}\text{the}\text{right}\Delta AYD,\\ & \sin 30^{\circ }=\frac{DY}{AD}\end{array}$$
$$\Rightarrow \frac{1}{2}=\frac{DY}{2}$$    (∵ Given that AD = 2)
$$\begin{array}{rl}& \Rightarrow DY=1.......\left(2\right)\\ & \cos 30^{\circ }=\frac{AY}{AD}\end{array}$$
$$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AY}{2}$$    (∵ Given that AD = 2)
$$\begin{array}{rl}& \Rightarrow AY=\sqrt{3}......\left(3\right)\\ & \text{From}\text{the}\text{right}\Delta CED,\\ & \tan 60^{\circ }=\frac{CE}{DE}\end{array}$$
$$\Rightarrow \tan 60^{\circ }=\frac{\left(CB-EB\right)}{YB}$$     [∵ CE = (CB - EB) and DE = YB)]
$$\Rightarrow \tan 60^{\circ }=\frac{\left(CB-DY\right)}{AB-AY}$$     [∵ EB = DY and YB = (AB - AY)]
$$\Rightarrow \tan 60^{\circ }=\frac{\left(x-1\right)}{\left(x-\sqrt{3}\right)}$$
[∵ CB = x, DY = 1(eq:2), AB = x(eq:1) and AY = $$\sqrt{3}$$ (eq:3)]
$$\begin{array}{rl}& \Rightarrow \sqrt{3}=\frac{\left(x-1\right)}{\left(x-\sqrt{3}\right)}\\ & \Rightarrow x\sqrt{3}-3=x-1\\ & \Rightarrow x\left(\sqrt{3}-1\right)=2\\ & \Rightarrow 0.73x=2\\ & \Rightarrow x=\frac{2}{0.73}=2.7\end{array}$$

i.e., the height of the mountain = 2.7 km