Question

A man is standing on the top of a multistorey building, which is 30m high,observes the angle of elevation of the top of a tower as 60 degree and the angle of depression of the based of the tower as 30 degree. Find the distance between the tower and building. Also find height of tower

Answer 1

Given height of the building AB = 30
Hence ED = 30 m
Let height of the tower be CD = h
Therefore, CE = h - 30
Let the horizontal distance between the tower and the building b e BD = x
Hence AE = x
In right ΔABD
tan30° = (AB/BD)
(1/√3) = 30/x
∴ x = 30√3 m
That is horizontal distance between the building and the tower is 30√3 m
In right ΔAEC
tan60° = (CE/AE)
√3 = (h - 30)/x
√3 = (h - 30)/(30√3)
90 = h - 30
∴ h = 120 m
Thus height of the tower is 120 m

Answer 2

Step-by-step explanation:

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