IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$ IN $ABC$ triangle $AB=28$, angle $C=120$, what is the value of smallest side of the triangle if $AC:BC=3:5$
I only got that smallest side should be $3x%$, how do i solve this?

The law of cosines gives
$$28^2=(3x)^2+(5x)^2-2(3x)(5x)\cos\frac{2\pi}3=34x^2+15x^2=49x^2$$
Thus $28=7x$, $x=4$ and the smallest side of the triangle is $3x=12$.
