A shooter is standing on the ground and is looking at a bird up in the sky, which is at a vertical distance of 11 m. The shooter points the gun towards the bird at an angle of elevation of 30° with the horizontal. If the shooter is holding the gun at a height of 1 m from the ground, then what is the horizontal distance between the bird and the shooter?

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$10 \sqrt{3} m$

$\frac{10}{\sqrt{3}} m$

10 m

$\sqrt{3} m$

answer is B.

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Detailed Solution

Let A be the position of the bird and B be the point at which the shooter is holding the gun. Let CD be ground level.

∴ BC = 1 m

⇒ AE = 11 m − 1 m = 10 m

In right ΔAEB,

$\tan 30 ° = \frac{A E}{E B} \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{E B} \Rightarrow E B = 10 \sqrt{3} m$

Thus, the horizontal distance between the bird and the shooter is $10 \sqrt{3} m$ .

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