Q)  An observer 1.5 m tall is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.

Ans: 

Step 1: Let’s draw a diagram for the given question:

Let the tower be AB and observer be CD. We need to find the angle of elevation ∠ BCE

Step 2: From the diagram, we can see that:

CD = AE = 1.5 m

Now since AB = AE + BE

∴ 30 = 1.5 + BE

∴ BE = 30 – 1.5 = 28.5

Step 3: In Δ  BCE, tan C =

∴ tan θ =

∴ tan θ = 1

Since, we know that tan 45⁰ = 1

∴ tan θ = 1 = tan 45⁰

∴ θ = 45⁰

Therefore, angle of elevation is 45⁰

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