Question

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 bridge is at the height of 30m from the banks, find the width of the river.

Answer 1

The width of the river is 3(1+√3) m.

Given,

The angles of depression of the banks on opposite side of the river are 30° and 45°

The bridge is at the height of 30m from the banks.

Consider the figure, while going through the following steps:

∠ QAP = ∠ PAD = 30°      (alternate angles)

∠ RPB = ∠ PBD = 45°       (alternate angles)

From figure it's clear that,

tan A = PD/AD

tan 30° = 3/AD

1/√3 = 3/AD

AD = 3√3 .........(1)

tan B = PD/BD

tan 45° = 3/BD

1 = 3/BD

BD = 3 .........(2)

Adding (1) and (2), we get

AD + BD = 3√3 + 3

⇒ AB = 3(1+√3)

Therefore, the width of the river = 3(1+√3) m

Answer 2

Let P be the point of observation on the Bridge and XPY be a horizontal line through O.

Let A and B be the points on the banks on the opposite sides or the river.

Then,

PQ = Height of bridge = 3m

$\sf{\angle{PAQ}=\angle{APX}=30°}$

And $\sf{\angle{PBQ}=\angle{BPY}=45°}$

From right ∆PQA, we have

$\sf\red{\tan30° = \frac{PQ}{AQ}}$

=$\sf\red{{\frac{1}{ \sqrt{3} } \frac{3}{AQ}}$

=$\sf\red{AQ = 3 \sqrt{3} m}$[/tex]

From right ∆PQB, we have

=$\sf\green{\tan45° = \frac{PQ}{QB}}$

=$\sf\green{{1={ \frac{3}{QB}}$

=$\sf\green{QB = 3m}$

Hence, the width of the river is

$\longrightarrow$$\sf\blue{AB=AQ+QB}$

$\longrightarrow$$\sf\blue{3√3+3}$

$\longrightarrow$$\sf\blue{3(√3+1)m}$