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Question
The angle of elevation of a tower from a point on the same level as the foot of the tower is 30°. On advancing 150 metres towards the foot of the tower, the angle of elevation of the tower becomes 60°. Show that the height of the tower is 129.9 metres (Use `sqrt3 = 1.732`)
Solution
Let h be the height of tower and angle of elevation of the foot of the tower is 30°, on advancing 150 m towards the foot of tower then an angle of elevation becomes 60°.
We assume that BC = x and CD = 150 m.
Now we have to prove height of the tower is 129.9 m.
So we use trigonometrical ratios.
In a triangle ABC
`=> tan C = (AB)/(BC)`
`=> tan 60^@ = (AB)/(BC)`
`=> sqrt3 = h/x`
`=> h/sqrt3 = x`
Again in a triangle ABD
`=> tan D = (AB)/(BC + CD)`
`=> tan 30^@ = h/(x + 150)`
`=> 1/sqrt3 = h/(x + 150)`
`=> x + 150 = sqrt3h`
`=> x = sqrt3h - 150`
`=> h/sqrt3 = sqrt3 - 150`
`=> h = 3h - 150sqrt3`
`=> 2h = 150sqrt3`
`=> h = (150 xx 1.732)/2`
=> h = 129.9
Hence the height of tower is 129.9 m proved
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