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Question
The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.
Solution
Let the height of tower be h meter and length of shawdow y meter initially.
In ΔABC,
tan 45° = `"AB"/"BC"`
1 = `h/y`
y = h ...(1)
In ΔABD,
tan 30° = `"AB"/"DB"`
`1/sqrt3 = h/(y + 10)`
y + 10 = h`sqrt3` ...(2)
Put y = h in equation (ii),
h + 10 = `hsqrt3`
`h(sqrt3 - 1) = 10`
h = `10(sqrt3 + 1)/((sqrt3 - 1)(sqrt3 + 1))`
h = `10/(3 - 1) (sqrt3 + 1)`
h = `10/2(sqrt3 + 1)`
h = 5(1.732 + 1)
h = 5 × 2.732
h = 13.66 meter
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