Question

Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from ${30}^0$ to ${45}^0$.

Answer 1

Let's assume $AB$ to be the building of height $hm$.
Let the two points be $C$ and $D$ be such that $CD=40m$.
In $△ABC$,
$\frac{AB}{BC}=tan{45}^0=1$
$BC=AB=h$
And, in $△ABD$,
$\frac{AB}{BD}=tan{30}^0$
$\sqrt{3}h=40+h$
$h=\frac{40}{0.732}=54.64m$
Therefore, the height of the building is $54.64m$.