From a point, 40 m apart from the foot of a tower, the angle of elevation of its top is 60°. The height of the tower is
AB is the tower = $$h$$ = ?
In $$\triangle$$ ABC,
=> $$tan(60^\circ)=\frac{AB}{BC}$$
=> $$\sqrt{3}=\frac{h}{40}$$
=> $$h=40\sqrt{3}$$ m
=> Ans - (C)
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