The angle of elevation ofthe top of a tree from a point on the ground which is 300 m away from thetree is $$30^\circ$$. When the tree grew up, its angle ofelevation ofthe top of it became $$60^\circ$$ from the same point. How much did the tree grow? (nearest to an integer)

we draw the diagram is given below

Ab is the height that tree grew up 

Let AB= h 

In $$\triangle BCD $$

$$\frac{BC} {CD} = \tan \30^\circ $$

BC =$$ 300 \times \frac{1}{\sqrt {3}} $$ 

BC = $$100 \sqrt {3}$$ m  

In $$ \triangle ACD $$

$$ \frac{AC} {CD} = \tan 60^\circ $$

$$ \Rightarrow \frac{AB+BC} {300} = \sqrt {3} $$

$$ AB + 100 \sqrt {3} = 300 \sqrt {3} $$

$$ AB = 300 \sqrt{3} - 100 \sqrt{3} $$

$$ AB = 200 \sqrt{3} $$ 

$$ AB = 200 \times 1.73 $$

$$ AB = 346 $$ 

therefore the tree grow = 346 m Ans