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Question
A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45°, (ii) 60°. Find the height of the tower in each case.
Solution
Let the height of the tower be h m.
i. Here, θ = 45°
∴ `tan 45^circ = (h - 1.6 )/20`
`=> 1 = (h - 1.6)/20`
∴ h = 21.6 m
So, height of the tower is 21.6 m
ii. Here θ = 60°
∴ `tan 60^circ = (h - 1.6)/20`
`=> sqrt(3) = (h - 1.6)/20`
∴ h = 20 × 1.732 + 1.6 = 36.24 m
So, height of the tower is 36.24 m.
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