From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5 m from the banks, find the width of the river. [Take √3 = 1.732.
In the above figure, let XY be the bridge and A be the point on the bridge from which two points, say B and D, on opposite sides of the river are observed. Join B and C. Given that the angles of depression of B and C from the point A are 30° and 45° respectively. Join B, D to A. So, ∠XAB = ∠ABD = 30° and also ∠YAD = ∠ADB = 45°. Again, draw a line AC from A perpendicular to the ground. Then, we get two right-angled triangles ABC and ACD with right angles at C. Now, given that the height of the bridge is AC = 2.5 m. We have to find the width of the river, that is BD.
From ∆ACD,
CD = 2.5m.
Again from ∆ABC,
or,
Hence, the width of the river is BD = BC + CD = 4.33 + 2.5 = 6.83m, which is the required solution.