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Question
The length of the shadow of a tower standing on the level plane is found to 2x meter longer when the sun's altitude is 30° than when it was 45°. Prove that the height of the tower is `x(sqrt3 + 1)` meters.
Solution
Let AB be the tower of height hm. the length of the shadow of the tower to be found 2xmeters at the plane longer when sun’s altitude is 30° than when it was 45°.
Let BC = ym,
CD = 2x m and ∠ADB = 30°, ∠ACB = 45°
We have to find the height of the tower
`We have the corresponding figure as follows
So we use trigonometric ratios.
In a triangle ABC
`=> tan C = (AB)/(BC)`
`=> tan 45^@ = h/y`
`=> 1 = h/y`
`=> y = h`
Again in a triangle ADB
`=> tan D = (AB)/(BC)`
`=> tan 45^@ = h/y`
`=> tan 30^@ = h/(2x + y)`
`=> 1/sqrt3 = h/(2x + y)`
`=> sqrt3h = 2x + y`
`=> h(sqrt3 - 1) = 2x`
`=> h = (2x)/(sqrt3 - 1)`
`=> h = x(sqrt3 + `)`
Hence the height of tower is `x(sqrt3 + 1)` m
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