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Question
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is 60° and the angle of elevation of the top of the pole, as seen from the foot of the tower is 30°. Find:
- the height of the tower;
- the horizontal distance between the pole and the tower.
Solution
Let AB be the tower and CD be the pole.
Given, CD = 20 m, ∠ADB = 60° and ∠CBD = 30°
In ΔBDC,
`(CD)/(BD) = tan 30^circ`
`=> BD = 20sqrt(3) m`
In ΔDBA,
`(AB)/(BD) = tan 60^circ = sqrt(3)`
`=> AB = 20sqrt(3) xx sqrt(3) = 60 m`
Hence,
i. Height of the tower = 60 m
ii. Horizontal distance between the pole and tower
= 20 × 1.732
= 34.64 m
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