Please Help

we know that one angle is \(30^\circ\) and another is \(90^\circ\) and the sum of all the angles in a triangle is \(180^\circ\) so we know that the third angle = 180 - 30 - 90 = \(60^\circ\) and so the triangle is a 30-60-90 degree triangle and since \(\overline{CA}\)= 6 then \(\overline{CB} = \overline{CA} \cdot 2=12 ~ and~ \overline{AB} = \overline{CA}\sqrt{3} = 6\sqrt3\) and the area is \(\frac{6\cdot6\sqrt{3}}{2}=\frac{6\cdot6\cdot\sqrt{3}}{2}=\frac{36\sqrt3}{2}=\boxed{18\sqrt3}\)

so since we know that this is a 30-60-90 triangle the base is 6sqrt3 and we can find that the area is 6*6sqrt3=36sqrt3 we then fivide by 2 to get 18sqrt3

The area of the triangle is \(\Huge{18\sqrt3}\)