I am going to draw a picture to ilustrate the solution
Now, we are going to use the trigonometric functions for the right triangle on the left to calculate h
[tex]\sin (30)\text{ = }\frac{h}{10\sqrt[]{3}}\Rightarrow\text{ h= sin(30)}\cdot\text{ 10}\sqrt[]{3\text{ }}=\frac{1}{2}\cdot10\sqrt[]{3}=5\sqrt[]{3}[/tex]
Now we use the formula for the area of a triangle
[tex]A=\frac{b\cdot h}{2}=\text{ }\frac{20\cdot5\sqrt[]{3}}{2}=50\sqrt[]{3}[/tex]
