Question 106
The angle of depression of a point from the top of a 200 m high tower is 45°. The distance of the point from the tower is
Solution
AB = tower = 200 m
$$\angle$$DAC = $$\angle$$ACB = 45
From $$\triangle$$ABC
=> $$tan 45 = \frac{AB}{BC}$$
=> $$1 = \frac{200}{BC}$$
=> $$BC = 200 m$$
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