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An aeroplane is flying at a height of $300 m$ above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are $45 °$ and $60 °$ respectively, then the width of the river is [Use $3 = 1 ⋅ 732$

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473.2 m

475.2 m

453.2 m

573.2 m

answer is A.

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Detailed Solution

Given that an aeroplane is flying at a height of

300 m

above the ground.
Consider the diagram,
Question Image

We know,

tanθ = \frac{Perpendicular}{Base}

\tan 60 ° = \sqrt{3}

\tan 45 ° = 1

Now in triangle DAC,

\tan 45 ° = \frac{DC}{CB}

\Rightarrow \tan 45 ° = \frac{300}{CB}

\Rightarrow CB = \frac{300}{\tan 45 °}

In triangle DBC,

\tan 60 ° = \frac{DC}{AB}

\Rightarrow AB = \frac{300}{\tan 60 °}

The width of the river is,
AB+CB
Substituting the values,

x = 300 tan 45 ° + 300 tan 60 ° ⇒ x = 300 + 3003 ⇒ x = 300 + 1003 ⇒ x = 473.2 m

The width of the river is 473.2 m.
Therefore, option 1 is correct.

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