An observer standing on a 300 m high tower observes two boats in the same direction; their angle of depression are 60 and 30 respectively. The distance between the boats is

173.2 m

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346.4 m

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25 m

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72 m

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Solution

The correct option is A 346.4 m

(Refer to Image)
Height of tower= $A B = 300 m$
Let distance between boats= $D C = y$ $m$
In $△^{l e} A B C, tan 60^{o} = \frac{300}{x}$
$\Rightarrow x = \frac{300}{\sqrt{3}}$
$\Rightarrow 100 \sqrt{3} m$
In $△ A B D, tan 30^{o} = \frac{300}{x + y}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{300}{x + y}$
$\Rightarrow 100 \sqrt{3} + y = 300 \sqrt{3}$
$\Rightarrow y = \sqrt{3} (300 - 100) = 200 \sqrt{3} m = 346.4 m$

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An observer standing on a 300 m high tower observes two boats in the same direction; their angle of depression are 60 and 30 respectively. The distance between the boats is

The correct option is A 346.4 m(Refer to Image)Height of tower= AB=300mLet distance between boats= DC=y mIn △leABC,tan60o=300x⇒x=300√3⇒100√3mIn △ABD,tan30o=300x+y⇒1√3=300x+y⇒100√3+y=300√3⇒y=√3(300−100)=200√3m=346.4m