In the diagram below, we have $AB = 24$ and $\angle ADB =90^\circ$. If $\sin A = \frac13$ and $\sin C = \frac12$, then what is $BC$?
sinA = 1/3
So
BD / AB = 1/3
BD / 24 = 1/3
BD = 24 (1/3) = 8
sin C = 1/2
Angle C = arcsin (1/2) = 30°
Then triangle BDC is a 30-60-90 right triangle
BD = 8
So BC = 2BD = 2 * 8 = 16
+730 
