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Question
The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60° . At a point Y, 40m vertically above X, the angle of elevation is 45° . Find the height of tower PQ.
Solution
We have
XY = 40m,∠PXQ = 60° and ∠MYQ = 45°
Let PQ = h
Also, MP = XY = 40m, MQ = PQ - MP = h - 40
In ΔMYQ,
` tan 45° = (MQ)/(MY)`
`⇒1 = (h-40)/(MY)`
⇒ MY = H - 40
⇒ PX = MY = h - 40 ................(1)
Now , in ΔMXQ,
`tan 60° = (PQ)/(PX)`
`⇒ sqrt(3) = h/( h-40)` [From (i)]
`⇒ h sqrt(3) - 40 sqrt(3) = h `
`⇒ h sqrt(3) -h = 40 sqrt(3)`
`⇒ h (sqrt(3)-1) = 40 sqrt(3)`
`⇒ h = (40sqrt(3))/((sqrt(3)-1))`
`⇒ h = (40 sqrt(3))/((sqrt (3)-1)) xx ((sqrt(3)+1))/((sqrt(3)+1))`
`⇒ h = (40 sqrt(3)(sqrt(3)+1))/((3-1))`
`⇒h = (40 sqrt(3) ( sqrt(3) +1))/2`
`⇒h= 20sqrt(3) (sqrt(3)+1)`
`⇒ h= 60+20 sqrt(3)`
`⇒ h= 60+20xx 1.73`
`⇒h = 60+ 34.6`
∴ h = 94.6m
So, the height of the tower PQ is 94. 6 m.
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