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Question
The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m. longer than when it was 600. Find the height of the tower.
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Solution
Let h be the height of tower AB and angle of elevation are 45° and 60° are given.
In a triangle OAC, given that AB = 10+x and BC = x
Now we have to find the height of the tower.
So we use trigonometrical ratios.
In a triangle OAB,
`=> tan A = (OB)/(AB)`
`=> tan 45^@ = (OB)/(AB)`
`=> 1= h/(10 + x)`
=> h = 10 + x
Therefore x = h - 10
Again in a triangle OCB
`=> tan C = (OB)/(BC)`
`=> tan 60^@ = (OB)/(BC)`
`=> sqrt3 = h/x`
`=> h = sqrt3x`
Put x = h - 10
`=> h = sqrt3 (h - 10)`
`=> h = sqrt3h - 10sqrt3`
`=> 10sqrt3 = h(sqrt3 - 1)`
`=> h = (10sqrt3)/(sqrt3 - 1)`
`=> h = (10 xx 1.732)/(1.732 - 1)`
`=> h = 17.32/0.327`
=> h = 23.66
Hence height of tower is 23.66 m
