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An airplane at an altitude of $200$ m observes the angle of depression of two opposite points on the two banks of a river to be $45^{o}$ and $60^{o}$ . Then, the width of the river is

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\frac{400}{\sqrt{3}}

400

\sqrt{3}

(200 - \frac{200}{\sqrt{3}})

(200 + \frac{200}{\sqrt{3}})

answer is D.

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Detailed Solution

It is given that an aeroplane at an altitude of

200

m observes the angle of depression of two opposite points on the two banks of a river to be

45^{o}

and

60^{o}

We have to find the width of the river.
We use the trigonometric identity

tanθ = \frac{Height}{Base}

.
From the given figure, A is the aeroplane, and B and C are the two points of the two banks of the river.
AD = height of aeroplane's altitude

= 200

∵ XAB = 45^{o}

∴ ∠ ABD = 45^{o}

{as they are alternate interior angles}
Similarily,

∵ ∠

YAC =

60^{o}

∴

∠

ACD =

60^{o}

Now, in the right-angled triangle ABD,

\tan 45^{o} = \frac{AD}{BD}

\Rightarrow 1 = \frac{200}{BD}

[

∵ \tan 45^{o} = 1

]

∴ BD = 200

....(1)
Also, in the right-angled triangle ADC,

\tan 60^{o} = \frac{AD}{DC}

\Rightarrow \sqrt{3} = \frac{200}{DC}

[

∵ \tan 60^{o} = \sqrt{3}]

∴ DC = \frac{200}{\sqrt{3}}

....(2) Now as we know,
Width of the river

= BD + DC

From

(1) and (2),

BD + DC

= (200 + \frac{200}{\sqrt{3}})

m
Hence, the width of the river is

(200 + \frac{200}{\sqrt{3}})

m.
Therefore, the correct option is 4.

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