Question

Hello Friends ❤❤❤❤❤❤❤❤

an observer 1.5m tall is 20.5 m away from a tower 22 m high.

determine the angle of elevation of the top of a tower from the eye of the observer.?????

Answer 1

$\textbf {Hey there, here is the solution.}$

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Horizontal distance from the observer to the tower = 20.5m (Given)

STEP 1: Find the vertical distance:

Vertical distance from the observer's eye to the height of the tower

= 22 - 1.5

= 20.5 m

STEP 2: Find the angle of elevation:

tan(θ) = opp/adj

tan(θ) = 20.5/20.5

tan(θ) = 1

θ = tan⁻¹ (1)

θ = 45°

Answer: The angle of elevation is 45°

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⭐$\textbf {Cheers}$⭐

Answer 2

Let the angle of elevation be A.

= > FC = BC - AE

= > FC = 22 - 1.5

= > FC = 20.5m.

Given EF = AB = 20.5m

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In angle, EFC

= > tanA = FC/EF

= > tanA = 20.5/20.5

= > tanA = 1

= > A = 45.

Therefore, the angle of elevation of the top of tower = 45.

Hope it helps!