# How do you find the other two side of a right triangle ABC, if ∠B=60 and AB=12 where AB is the hypotenuse?

Nov 6, 2014

$m a t h b f \left\{{30}^{\circ} \text{-"60^circ"-} {90}^{\circ}\right\}$ Triangle

The ratios of three sides of a ${30}^{\circ} \text{-"60^circ"-} {90}^{\circ}$ triangle are:

$1 : \sqrt{3} : 2$

Since the posted Triangle ABC is a ${30}^{\circ} \text{-"60^circ"-} {90}^{\circ}$ triangle, we have the ratios:

$B C : A C : A B = 1 : \sqrt{3} : 2$

$A B = 12 \implies B C = \frac{A B}{2} = \frac{12}{2} = 6$

$A C = B C \cdot \sqrt{3} = 6 \sqrt{3}$

Hence, $B C = 6$ and $A C = 6 \sqrt{3}$.

I hope that this was helpful.