Question
A sniper on the parapet of a 50 feet high building was aiming at a target. The angle of depression that his rifle made was 45 °. How far away from the building was his target?
Let θ be the angle of depression of the point on the ground as seen from the top of a tower, here θ = 45° Let AC be the height of the tower, here AC = 50 feet. Let the distance of the point on the ground from the foot of the tower, AB = x feet. Here, tan θ = AC/AB ⇒ tan 45° = 50/x ⇒ 1 = 50/x ⇒ x = 50 feet More Height and Distance Questions
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