From a point on a bridge across a river, the angles of depression of the banks on opposite side of the river are 30° and 45° respectively. If the bridge is at the height of 30 m from the banks, find the width of the river.
Let BD be the width of the river. And the angle of depression of the bank on opposite side of the river is 30° and 45° respectively. It is given that AC = 30 m.
Let BC = x and CD = y. And ∠ABC = 30°, ∠ADC = 45°.
Here we have to find the width of the river.
We have the following figure
So we use trigonometric ratios.
In a triangle ABC
`=> tan 30° = (AC)/(BC)`
`=> 1/sqrt3 = 30/x`
`=> x = 30sqrt3`
Again in a triangle ADC
`=> tan 45° = (AC)/(CD)`
`=> 1 = 30/y`
`=> y = 30`
So width of river is
`x + y = 30sqrt3`
`x + y = 30(sqrt3 + 1)`
Hence the width of river is `(30(sqrt3 + 1)`m