A boy, standing on a 50 m tall building is looking at a stone on the ground. The distance between the stone and the building is equal to the height of the building. Find the angle of depression.

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Solution

From the figure, CA = 50 m
Given : AB = CA = 50 m
To find: The angle of depression $∠ D C B$

DC and AB are the horizontals.
$∠ C D B = ∠ A B C = θ$ (Alternate interior angles)
So, in $△ A B C$ ,
$t a n θ = \frac{A C}{A B} = \frac{50}{50} = 1$
We know that, $t a n θ = 1,$ when, $θ = 45^{o}$
$\Rightarrow ∠ A B C = ∠ B C D = 45^{o}$
Therefore, angle of depression = $45^{o}$

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A boy, standing on a 50 m tall building is looking at a stone on the ground. The distance between the stone and the building is equal to the height of the building. Find the angle of depression.

From the figure, CA = 50 m Given : AB = CA = 50 m To find: The angle of depression ∠DCB DC and AB are the horizontals. ∠CDB=∠ABC =θ (Alternate interior angles) So, in △ABC, tan θ=ACAB=5050=1 We know that, tan θ=1, when, θ=45o ⇒∠ABC=∠BCD=45o Therefore, angle of depression = 45o