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Question
In the given figure, from the top of a building AB = 60 m hight, the angle of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find:
- the horizontal distance between AB and CD.
- the height of the lamp post.
Solution
Given that AB is a building that is 60 m, high.
Let BC = DE = x and CD = BE = y
`=>` AE = AB – BE = 60 – y
i. In right ΔAED,
`tan 30^circ = (AE)/(DE)`
`=> 1/sqrt(3) = (60 - y)/(x)`
`=> x = 60sqrt(3) - ysqrt(3)` ...(1)
In right ΔABC,
`=> tan 60^circ = (AB)/(BC)`
`=> sqrt(3) = 60/x`
`=> x = 60/sqrt(3)`
`=> x = 60/sqrt(3) xx sqrt(3)/sqrt(3)`
`=> x = (60sqrt(3))/3`
`=> x = 20sqrt(3)`
`=>` x = 20 × 1.732
`=>` x = 34.64 m
Thus, the horizontal distance between AB and CD is 34.64 m.
ii. From (1), we get the height of the lamp post = CD = y
`x = 60sqrt(3) - ysqrt(3)`
`=> 20sqrt(3) = 60sqrt(3) - ysqrt(3)`
`=>` 20 = 60 – y
`=>` y = 60 – 20
`=>` y = 40 m
Thus, the height of the lamp post is 40 m.
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