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From the top of a 120 m high tower, the angle of depression of the top of a pole 45° and the angle of depression of the foot of the pole is θ, such that tan θ = 3/2, what is the height of the pole?

From the top of a 120 m high tower, the angle of depression of the top of a pole 45° and the angle of depression of the foot of the pole is θ, such that tan θ = 3/2, what is the height of the pole? Correct Answer 40 m

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Given, tan θ = 3/2

From the following figure

Height of the Tower AB = 120 m

Height of the pole = EC

DE = BC and BD = EC

In ΔABC

tan θ = AB/BC

3/2 = 120/BC

BC = 80 cm

DE = BC = 80 cm

In ΔAED

tan 45° = AD/DE

1 = AD/DE

AD = 80 m

AB = BD + DA

120 = BD + 80

BD = 120 - 80 = 40 m

Hence, height of the pole is 40 m

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