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Question

The angles of depression of the top and bottom of a building 50 meters high as 4 observed from the top of a tower are 30 and 60, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

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Solution

In ΔBTP,

tan 30° = TP/BP

1/√3 = TP/BP

BP = TP√3

In ΔGTR,

tan 60° = TR/GR

√3 = TR/GR

GR = TR/√3

As BP = GR

TP√3 = TR/√3

3 TP = TP + PR

2 TP = BG

TP = 50/2 m = 25 m

Now, TR = TP + PR

TR = (25 + 50) m

Height of tower = TR = 75 m

Distance between building and tower = GR = TR/√3

GR = 75/√3 m = 25√3 m


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