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)