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Question
From the top of a vertical tower, the angles depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60° . If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.
Solution
Let OP be the tower and points A and B be the positions of the cars.
We have,
AB =100m, ∠OAP = 60° and ∠OBP = 45°
Let OP = h
In ΔAOP,
` tan 60° = (OP)/(OA)`
`⇒ sqrt(3) = h/(OA)`
`⇒OA = h/sqrt(3)`
Also, in Δ BOP ,
` tan 45° = (OP)/(OB)`
`⇒ 1 = h/(OB)`
⇒ OB = h
Now,OB - OA = 100
`⇒ h - h/sqrt(3) = 100`
` ⇒ (h sqrt(3)-h)/sqrt(3) = 100`
`⇒(h(sqrt(3) -h))/sqrt(3) = 100`
`h = (100sqrt(3) )/((sqrt(3)-1)) xx ((sqrt(3)+1))/((sqrt(3)+1))`
`h = (100 sqrt(3) ( sqrt(3)+1))/((3-1))`
`⇒ h = (100(3+sqrt(3)))/2`
⇒ h = 50(3+1.732)
⇒ h =50(4.732)
∴ h =236.6m
So, the height of the tower is 236.6 m.
Disclaimer. The answer given in the textbook is incorrect. The same has been rectified above.
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