# How do you solve the right triangle given the hypotenuse is 12 and it is a 30-60-90 triangle?

$\sin A = \sin 30 = \frac{1}{2} = \frac{a}{c} = \frac{a}{12} \to a = \frac{12}{2} = 6$
$\cos A = \cos 60 = \frac{\sqrt{3}}{2} = \frac{b}{12} - \to b = 6 \sqrt{3} = 10.39$