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An observer 1.75 m tall is at a distance of 24 m from a wall 25.75 m high. What will be the angle of elevation of the top of the wall at the observer's eye?

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45 °

25 °

65 °

40 °

answer is A.

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Detailed Solution

Given that,
The given height of the wall AE = 25.75 m
The height of a man BD = CE = 1.75 m
The distance between man and wall BC=DE= 24 m
The angle of elevation =

θ

We draw the diagram,
Question Image

For the height of the wall above the line of sight we substitute

H M

=1.75 for man-height and

H W

=25.75 for wall-height in the equation

H e i g h t = H w − H M

⇒ H e i g h t = 25.75 − 1.75 ⇒ H e i g h t = 24

Substituting height= 24, BC=24 for base and

θ

for angle in the equation

t a n θ = h e i g h t b a s e

⇒ t a n θ = 2424 ⇒ t a n θ = 1 ⇒ θ = t a n − 1 1 [∵ tan 45 ° = 1] ⇒ θ = t a n − 1 tan 45 ° ⇒ θ = 45 °

Thus the angle of elevation is

45 °

from the observer’s eye.
Therefore the correct option is 1.

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