Question 69

The height of a tower is 50√3 m. The angle of elevation of a tower from a distance 50 m from its feet is

Solution

Given : AB is the tower = $$50\sqrt{3}$$ m and BC = 50 m

To find : $$\angle$$ ACB = $$\theta$$ = ?

Solution : In $$\triangle$$ ABC

=> $$tan(\theta)=\frac{AB}{BC}$$

=> $$tan(\theta)=\frac{50\sqrt{3}}{50}$$

=> $$tan(\theta)=\sqrt{3}$$

=> $$\theta=tan^{-1}(\sqrt{3})$$

=> $$\theta=60^\circ$$

=> Ans - (C)

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