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Question
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man's eye is 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as x°, where tan x° = `2/5`.
Calculate :
- the distance AB in metres;
- angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.
Solution
i. Let AD be the height of the man, AD = 2 m
∴ CE = (10 – 2) = 8 m
In ΔCED,
`(CE)/(DE) = tan x^circ = 2/5`
`=> 8/(DE) = 2/5 `
`=>` DE = 20 m
Here AB = DE
∴ AB = 20 m
ii. Let AD be the height of the man, AD = 2 m.
∴ CE = (10 – 2) = 8 m
Let A"D" be the new position of the man and θ be angle of elevation of the top of the tower.
So, D'E = 15 m
In ΔCED,
`tan theta = (CE)/(D^’E)`
= ` 8/15`
= 0.533
`=>` θ = 28°
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