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Question
The horizontal distance between towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60m, find the height of the first tower.
Solution
Let AB and PQ be the two towers.
BQ = AR = 140 m
AB = RQ = 60 m
Let PR = h
We need to find the height of the first tower, i.e. , PQ.
In ΔPRA,
`"PR"/"AR" = tan30^circ`
`"h"/140 = 1/sqrt(3)`
`"h" = 140/sqrt(3) = 80.83`
∴ PQ = PR + RQ = 80.83 m + 60 m = 140.83 m
Thus , the height of the first tower is 140.83 m.
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