A tree breaks due to storm and the broken part bends so that the top of the tree first touches the ground, making an angle of 30 with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is 10 m. The height of the tree is

Given : BC = 10 m and $$\angle ACB=30^\circ$$

To find : AC + AB = ?

Solution : In right $$\triangle$$ ABC

=> $$tan(30^\circ)=\frac{AB}{BC}$$

=> $$\frac{1}{\sqrt3}=\frac{AB}{10}$$

=> $$AB=\frac{10}{\sqrt3}$$ ---------------(i)

Similarly, $$cos(30^\circ)=\frac{BC}{AC}$$

=> $$\frac{\sqrt3}{2}=\frac{10}{AC}$$

=> $$AC=\frac{20}{\sqrt3}$$ ------------------(ii)

$$\therefore$$ Height of tree = $$AB+AC$$

= $$\frac{10}{\sqrt3}+\frac{20}{\sqrt3}$$

= $$\frac{30}{\sqrt3}=10\sqrt3$$ m

=> Ans - (B)