Q)  An observer 1.5 m tall is 28.5 m away from a tower and the angle of elevation of the top of the tower from the eye of the observer is
45 degrees. What is the height of the tower?

Ans: 

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

Let the tower be AB and observer be CD. The angle of elevation from point C is 45 and distance between points A and D is 28.5 m.

We need to find the value of height of tower, H.

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

CD = AE = 1.5 m

Now since AB = AE + BE

∴ H = 1.5 + BE

∴ BE = H – 1.5

Step 3: In Δ  BCE, tan C =

∴ tan 45⁰ =

∴ 1 =

∴ H – 1.5 = 28.5

∴ H = 28.5 + 1.5

∴ H = 30

Therefore, the height of the tower is 30 m.

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