let $\theta$ be the top angle of the triangle and $\gamma$ be the angle opposite $R_b$
using the law of sines
$\dfrac{R_a}{\sin(m)} = \dfrac{C}{\sin(\theta)}$
$\sin(\theta) =\dfrac{C \sin(m)}{R_a}$
$\theta = \arcsin\left(\dfrac{C \sin(m)}{R_a}\right)$
$\gamma = \pi - m - \theta$
$\dfrac{R_b}{\sin(\gamma)}=\dfrac{R_a}{\sin(m)}$
$R_b = \dfrac{R_a \sin(\gamma)}{\sin(m)}$