Question

The height of a tower is 1003. Find the angle of elevation of its top from a point 100 mt away from
its foot.

Answer 1

Correct Question:

The height of a tower is 100√3. Find the angle of elevation of its top from a point 100 m away from its foot.

Answer:

30° is the angle of elevation.

Step-by-step explanation:

Let AC be the tower of height 100√3 m and B be the point at a distance of 100 m away from the foot of the tower.

Let θ be the angle of elevation from point B of top of the tower.

In ∆ABC,

• tan θ = AC/BC
• tan θ = 100√3/100
• tan θ = 100√3 × 1/100
• tan θ = √3
• tan θ = 60°

∴ The angle of elevation of its top from a point 100 m away from its foot = 60°

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