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Question
The angle of elevation of the top of a tower from ta point on the same level as the foot of the tower is 30° . On advancing 150 m towards foot of the tower, the angle of elevation becomes 60° Show that the height of the tower is 129.9 metres.
Solution
Let AB be the tower
We have:
CD =150m,∠ACB = 30° and ∠ADB = 60°
Let:
AB = hm and BD = xm
In the right ΔABD,we have:
`(AB)/(AD) = tan 60° = sqrt(3)
`⇒h/x = sqrt(3)`
`⇒ x = h/sqrt(3)`
Now, in the right ΔACB,we have:
`(AB)/(AC) = tan 30° = 1/ sqrt(3)`
`⇒ h/(x +150) = 1/ sqrt(3)`
`sqrt(3) h =x +150`
On putting `x= h/ sqrt(3) ` in the above equation, we get:
`sqrt(3)h = h/ sqrt(3) + 150`
`⇒ 3h = h + 150 sqrt(3)`
`⇒ 2h = 150sqrt(3)`
`⇒ h = (150 sqrt(3)) /2 =75 sqrt(3) = 75 xx1.732 = 129.9m`
Hence, the height of the tower is 129.9 m
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