# How do you solve the triangle if we are given a = 9, b = 12, c = 15?

Apr 23, 2015

You might recognize these proportions as being those of the standard $3 : 4 : 5$ right angled triangle

which immediately gives us
$\angle C = \frac{\pi}{2} \left(= {90}^{o}\right)$

$\sin \left(\angle A\right) = \frac{12}{15} = 0.8$
So
$\angle A = \arcsin \left(0.8\right)$

$\sin \left(\angle B\right) = \frac{9}{15} = 0.6$
So
$\angle B = \arcsin \left(0.6\right)$