NIMCET 2013 Question Paper with solutions

Question : 21 of 120

Marks: +1, -0

A man observes the angle of elevation of the top of a mountain to be 30°. He walks 1000 feet nearer and finds the angle of elevation to be 45°. What is the distance of the first point of observation from the foot of the mountain?

500 $√3$ ( $√3$ + 1) ft.
500 ( $√3$ + 1) ft.
500 ( $√3$ - 1) ft.
500 $√3$ ( $√3$ - 1) ft.

Validate

Solution:

As per the given information in the question, we can draw the following figure

where, AB = x ft
DC = 1000 ft
CB = x ft
In ΔABD,

{AB}/{BD}=\tan θ

⇒\tan 30°=x/{x+1000}

⇒1/√3=x/{x+1000}

⇒√3x=x+1000

⇒√3x-x=1000

⇒x=1000/{√3-1}

⇒x=1000/{√3-1}

∴BD=1000+1000/{√3-1}=500√3(√3+1)

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