A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC upto B, a distance of 100 m and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO, but he finds the line of sight of C shifted from the original line of sight by angle $θ = 40^{0}$ $(θ i s k n o w n a s p a r a l l a x),$ the distance of tower C from his original position A is

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19 m

119 m

100 m

10 m

answer is D.

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Detailed Solution

Given, parallax angle $θ = 40^{0}$

From the given figure, $A B = A C \tan θ$

$\Rightarrow A C = \frac{A B}{\tan θ} = \frac{100 m}{\tan 40^{0}} = \frac{100}{0.8391} = 119 m$

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