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Question
A man observes the angle of elevation of the top of the tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures.
Solution
Let the height of the tower be ‘h’ m.
In Δ ADC , tan 45° = `h/(20 + x)`
1 = `h /(20+ x`
⇒ h = 20 + x
Also , In ΔBDC , tan 60° = `h/x`
`sqrt(3) = h / x`
⇒ x = `h/(sqrt(3))` ...(2)
h = 20 + `h/sqrt(3)`
`h - h/sqrt(3) = 20`
`h((sqrt(3) -1)/sqrt(3)) = 20`
`h = (20 sqrt(3)) /((sqrt(3) - 1)) xx ((sqrt(3) + 1)) /(( sqrt(3) +1 )`
`= (20 (3 + sqrt(3)))/(3-1)`
`= (20 (3+1.732))/2`
= 10 (4.732)
Height of the tower . h = 47.32 m
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