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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.

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3 (\sqrt{3} - 1)

3 (\sqrt{3} + 1)

(3 + \sqrt{3})

(3 - \sqrt{3})

answer is A.

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Detailed Solution

It is given that the angles of depressions of the banks on opposite sides of the river are

30^{\circ}

and

45^{\circ}

The bridge is at a height of

3 m

from the banks
We have to find

tanθ = \frac{perpendicular}{base} .

Let us assume the width of river =AB
And bridge is at height of

3 m

from banks
Consequently,

DP = 3 m

So,

∠ QPA = 30^{0}

∠ RPB = 45^{0}

We can see that AC will be perpendicular to PD.
Also, line QR is parallel to line AB

∠ PAD = ∠ QPA = 30^{0}

(Alternate angle)
Now, in triangle PAD,

\tan 30^{0} = \frac{PD}{AD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{3}{AD}

[since

\tan 30^{0} = \frac{1}{\sqrt{3}}

]

AD = 3 \sqrt{3}

Now in triangle PBD ,

\tan 45^{0} = \frac{PD}{DB}

\Rightarrow 1 = \frac{3}{DB}

[since

\tan 45^{0} =

\Rightarrow DB = 3

The width will be,
AB=AD+DB

\Rightarrow

= 3 \sqrt{3} + 3

\Rightarrow AB = 3 (\sqrt{3} + 1)

Hence width of river is

3 (\sqrt{3} + 1) m

.
Therefore, the correct option is 1.

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