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Question
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second is 30° and 24° respectively. Find the height of the two towers. Give your answers correct to 3 significant figures.
Solution
In ΔAEC,
`tan 30^circ = (AE)/(EC)`
`=> 1/sqrt(3) = (AE)/120`
`=> AE = 120/sqrt(3)`
= `120/sqrt(3) xx sqrt(3)/sqrt(3)`
= `(120sqrt(3))/3`
= `40sqrt(3)`
= 40 × 1.732
= 69.28 m
In ΔBEC,
`tan 24^circ = (EB)/(EC)`
`=> 0.4452 = (EB)/120`
`=>` EB = 53.424 m
Thus, height of first tower,
AB = AE + EB
= 69.28 + 53.424
= 122.704
= 123 m ...(Correct to 3 significant figures)
And height of second tower,
CD = EB
= 53.424 m
= 53.4 m ...(Correct to 3 significant figures)
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