Question

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are ${30}^{\circ }and{45}^{\circ }$ respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river. [Take $\sqrt{3}=\mathrm{1.732.}$]

Answer 1

In $△$ ABD,

$tan{30}^o=\frac{2.5}{AB}\frac{1}{\sqrt{3}}=\frac{2.5}{AB}AB=2.5\times \sqrt{3}=2.5\times 1.732=4.33$

In $△$ BDC,

$tan{45}^o=\frac{2.5}{BC}1=\frac{2.5}{BC}BC=2.5$

Width of river = AB + BC

= 4.33 + 2.5 = 6.83 metre