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Question
The angle of elevation of the top of a tower is observed to be 60°. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45°. Find:
- the height of the tower,
- its horizontal distance from the points of observation.
Solution
In ΔABC
`tan 60^circ = (h + 30)/x`
`xsqrt(3) = h + 30` ...(1)
In ΔADE
`tan 45^circ = h/x`
`1 = h/x`
h = x ...(2)
From equation (1)
`xsqrt(3) = x + 30`
`x(sqrt(3) - 1) = 30`
`x = 30/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1)`
`x = (30(sqrt(3) + 1))/(3 - 1)`
= `(30(1.732 + 1))/2`
= 15 × 2.732
= 40.980 m
From equation (2)
h = 40.98 m
Height of tower = h + 30
= 40.98 + 30
= 70.98 m
= 71 m (approx.)
Horizontal distance = x = 40.98 m = 41 m.
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