Explanation:


Let A be the foot and C be the summit of a mountain.
Given that $\angle$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 $\angle$DAY = 30°

It is also given that from the point D, the elevation is 60°
i.e., $\angle$CDE = 60°

From the right $\triangle$ ABC,

tan45°=$\dfrac{CB}{AB}$

=>1=$\dfrac{x}{AB}$ [∵ CB = x(the height of the mountain)]

AB = x ⋯(1)

From the right $\triangle$ AYD,

sin30°=$\dfrac{DY}{AD}$

=>$\dfrac{1}{2}=\dfrac{DY}{2}$ (∵ Given that AD = 2)
=> DY = 1 ... (2)

cos30°=$\dfrac{AY}{AD}$
=> $\dfrac{\sqrt{3}}{2}=\dfrac{AY}{2}$ (∵ Given that AD = 2)
=> AY
=√3 ... (3)
From the right $\triangle$ CED,
tan60°=$\dfrac{CE}{DE}$
⇒tan60°=$\dfrac{CE}{DE}$
=>tan60°=$\dfrac{(CB-EB)}{YB}$ [∵ CE=(CB-EB) and DE=YB)]
=>tan60°=$\dfrac{(CB-DY)}{(AB-AY)}$ [∵ EB=DY and YB=(AB-AY)]
=>tan60°=$\dfrac{(x - 1)}{(x−\sqrt{3})}$ [∵ CB=x, DY=1(eq:2), AB=x(eq:1) and AY = √3(eq:3)]
⇒√3=$\dfrac{(x - 1)}{(x−\sqrt{3})}$
⇒x√3−3=x−1
⇒x(√3−1)=2
⇒0.73x=2
⇒x=$\dfrac{2}{0.73}$=2.7
i.e., the height of the mountain = 2.7 km