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Question
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of the tower is `y(sqrt(3) + 1)` metres.
Solution
Let AB be the tower and AB = x
Distance CD = 2y
In right ΔADB, we have
`tan theta = (AB)/(DB)`
`\implies tan 45^circ = x/(DB)`
`\implies 1 = x/(DB)`
`\implies` DB = x.
In right ΔACB, we have
`tan 30^circ = (AB)/(CB)`
`\implies 1/sqrt(3) = x/(CB)`
`\implies CB = sqrt(3)x`
`\implies x + 2y = sqrt(3)x`
∴ `sqrt(3)x - x = 2y`
`\implies x(sqrt(3) - 1) = 2y`
`x = (2y)/(sqrt(3) - 1)`
= `(2y(sqrt(3) + 1))/((sqrt(3) - 1)(sqrt(3) + 1))`
= `(2y(sqrt(3) + 1))/(3 - 1)`
= `(2y(sqrt(3) + 1))/2`
= `y(sqrt(3) + 1)`
∴ Required height of tower = `y(sqrt(3) + 1)`
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