I need help figuring out how to find angle B and side b and c?

Given:
A = 30°, C = 102°, a = 20;
The sum of all internal angles of a triangle is 180°; so:
B = 180° - 30° - 102°
B = 48°
Now, using the sine law:
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{c}{\sin C} \\ \\ \frac{20}{\sin30\text{\degree}}\text{ = }\frac{c}{\sin 102\text{\degree}} \\ \\ c\text{ = }\frac{\sin 102\text{\degree{}x 20}}{\sin \text{ 30\degree}}\text{ = 39}.1 \end{gathered}[/tex]
Finally:
[tex]\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B} \\ \\ \frac{20}{\sin30\text{\degree}}\text{ = }\frac{b}{\sin 48\text{\degree}} \\ \\ b\text{ = }\frac{20\text{ x sin 48\degree}}{\sin \text{ 30\degree}}\text{ = 29.7} \end{gathered}[/tex]