NIMCET 2025 Paper

Use $\tan =$ opposite $∕$ adjacent .
The angle subtended by the tower gives a relation between height and distance.
The depression angle from a higher point gives another relation.
Solve the two equations to find the horizontal distance.
Solution
Let the horizontal distance be $\mathit{d}$.
From ground point:
$\tan {30}^{\circ }=\frac{\mathit{H}}{\mathit{d}}\Rightarrow \mathit{H}=\frac{\mathit{d}}{\sqrt{3}}$
From point at height $\mathit{h}$ :
$\tan {60}^{\circ }=\frac{\mathit{h}}{\mathit{d}}\Rightarrow \mathit{d}=\frac{\mathit{h}}{\sqrt{3}}$
Thus,
$\mathit{d}=\mathit{h}\cot {60}^{\circ }$