Question

From the point on a bridge across a river, the angles of depressions of the banks on opposite sides of the river are $${30}^{\circ }$$ and $${45}^{\circ }$$, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Answer 1

Hint: Consider the maximum height as AB and draw the angle of depression at two different points and apply $$\tan \theta$$ to the two right angled triangles and we will get two equations and then we have to compute the maximum height from which he falls.

Complete step-by-step answer:

Let width of river =AB

And bridge is at height of 3m from banks So, DP=3m
Angel of depression of banks on the opposite sides of the river are $${30}^{\circ }$$ and $${45}^{\circ }$$.
$$\begin{array}{rl}& \mathrm{\angle }QPA={30}^{\circ }\\ & \mathrm{\angle }RPB={45}^{\circ }\end{array}$$

Now in triangle PAD

$$\tan {30}^{\circ }=PDAD$$

$$1\sqrt{3}=3AD$$

$$AD=3\sqrt{3}$$. . . . . . . . . . . . . . . . . . . . . . . (1)

Now in triangle PBD
$$\tan {45}^{\circ }=PDDB$$

$$1=3DB$$

$$DB=3$$. . . . . . . . . . . . . . . . . . . . . . . . . . (2)

Width of river = $$AB=AD+DB$$
Hence width of river $$=3(\sqrt{3}+1)$$

Note: If the object observed by the observer is below the level of the observer, then the angle formed between the horizontal line and the observer’s line of sight is called the angle of depression. As the person moves from one point to another angle of depression varies.