Answer
$sin~C = 1$
$C = 90^{\circ}$
Triangle ABC is a $30^{\circ},60^{\circ},90^{\circ}$ triangle.
Work Step by Step
We can use the law of sines to find the angle $C$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$sin~C = \frac{c~sin~A}{a}$
$sin~C = \frac{(2\sqrt{5})~sin~(30^{\circ})}{\sqrt{5}}$
$sin~C = 1$
$C = arcsin(1)$
$C = 90^{\circ}$
Triangle ABC is a $30^{\circ},60^{\circ},90^{\circ}$ triangle.
