Question 113
From the top of a tower 100m high, a person observes that the angle of elevation of the top of another tower is 60° and the angle of depression of the bottom of the tower is $$30^{\circ}$$. Then the height (in meters) of the second tower approximately is.
Solution
From A, angle of elevation to C i.e. angle CAX is 60 degrees and from A, angle of depression to D i.e. XAD is 30 degrees.
As AX is parallel to BD, angle BDA = XAD = 30 degrees.
Tan BDA = AB / BD where AB = 100 m.
This gives BD = $$100\sqrt{\ 3}$$ m. AX = BD = $$100\sqrt{\ 3}$$ m.
In triangle CAX, Tan CAX = CX/AX.
This gives CX = 300 m and length of CD becomes XD + CX = AB + CX = 400 m.
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