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Question
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°. Find:
- the height of the tower, if the height of the pole is 20 m;
- the height of the pole, if the height of the tower is 75 m.
Solution
Let AB be the tower and CD be the pole.
Then ∠ACE = 60° and ∠BCE = 30°.
i. In ΔBEC,
`(BE)/(EC) = tan 30^circ`
`=> 20/(EC) = 1/sqrt(3)`
`=> EC = 20sqrt(3) m`
In ΔAEC,
`(AE)/(EC) = tan 60^circ`
`=> AE = 20sqrt(3) xx sqrt(3) = 60 m `
∴ Height of the tower = AB
= AE + EB
= (60 + 20)
= 80 m
ii. Let height of the pole be x m
∴ CD = BE = x
In ΔBEC,
`(BE)/(EC) = tan 30^circ`
`=> EC = sqrt(3)x`
In ΔAEC,
`(AE)/(EC) = tan 60^circ`
`=> (75 - x)/(EC) = sqrt(3)`
`=>` 75 – x = 3x
∴ `x = 75/4 = 18.75 m`
∴ Height of the pole is 18.75 m.
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