A boat is on the one side of a bridge and the other boat is on the other side of the bridge. The bridge is 36 meter above the river surface. The angles of depression of both the boats from the bridge point on the bridge are complementary to each other and the angle of depression of first boat from the bridge is tan-1(1/3), then what is the distance between both the boats?

A boat is on the one side of a bridge and the other boat is on the other side of the bridge. The bridge is 36 meter above the river surface. The angles of depression of both the boats from the bridge point on the bridge are complementary to each other and the angle of depression of first boat from the bridge is tan-1(1/3), then what is the distance between both the boats? Correct Answer 120 meters

GIVEN:

The bridge is 36 meter above the river surface.

Angle of elevation of first boat from the bridge is Tan^-1(1/3)

CONCEPT:

Application of trigonometric ratios.

Sum of complementary angles is 90°

FORMULA USED:

tan A = Perpendicular/Base

CALCULATION:

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Let the distance of boat C from point A be x

And the distance of boat D from the point A be y

Let the angle of depression of boat C from the bridge be θ

So, the angle of depression of boat D from the bridge be (90 – θ)

Given that,

θ = tan^-1(1/3)

⇒ tan θ = 1/3

Now,

⇒ tan θ = AB/AC = 36/x

⇒ 1/3 = 36/x

⇒ x = 108 meters

And

tan (90 – θ) = AB/AD

⇒ tan (90 – θ) = 36/y

⇒ cot θ = 36/y

⇒ 3 = 36/y

⇒ y = 12 meters

⇒ x + y = 108 + 12 = 120 meters

∴ Distance between the boats is 120 m