Question:
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degree and the angle of depression of its foot is 45 degree . Determine the height of the tower
Answer:
Let BD = 7 is height of the building and AE is the height of the tower.
From the figure BD = CE = 7
Now from Δ BED
tan 45 = BD/DE
=> 1 = 7/DE
=> DE =7
Now DE = BC = 7
Again in Δ ABC
tan 60 = AC/BC
=> √3 = AC/7
=> AC = 7√3
The height of the tower = AE = AC + CE
= 7√3 + 7
= 7(√3 + 1) m
So height of the tower is 7(√3 + 1) m
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