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Question 80
From the top of a tower, the angles of depression of two objects on the ground on the same side of it, are observed to be $$60^\circ$$ and $$30^\circ$$ respectively and the distance between the objects is $$400 \surd3$$ m. The height (in m) of the tower is:
Solution
BC =Â $$400 \surd3$$ m
In $$\triangle$$ ACD,
$$tan 60\degree = \frac{\sqrt{3}}{1} = \frac{AD}{CD}$$Â
In $$\triangle$$ ABD,
$$tan 30\degree = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} =Â \frac{AD}{BD}$$
BC = BD - CD = 3 - 1 = 2 units
2 units =Â $$400 \surd3$$
$$\sqrt{3} units = \frac{400 \surd3}{2} \times \sqrt{3} = 600 m
The height of the tower = 600 m.
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