To find the length of side \( b \) in triangle \( ABC \), we can use the Law of Sines, which states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Given:
- \( a = 10 \) (side opposite angle \( A \))
- \( \angle A = 30^\circ \)
- \( \angle B = 45^\circ \)
We need to find \( \sin A \) and \( \sin B \):
\[ \sin A = \sin 30^\circ = \frac{1}{2} \] \[ \sin B = \sin 45^\circ = \frac{\sqrt{2}}{2} \]
Now we can set up the equation using the Law of Sines:
\[ \frac{10}{\frac{1}{2}} = \frac{b}{\frac{\sqrt{2}}{2}} \]
Calculating the left side:
\[ \frac{10}{\frac{1}{2}} = 10 \times 2 = 20 \]
So we have:
\[ 20 = \frac{b}{\frac{\sqrt{2}}{2}} \]
To solve for \( b \), multiply both sides by \( \frac{\sqrt{2}}{2} \):
\[ b = 20 \times \frac{\sqrt{2}}{2} \]
Calculating \( b \):
\[ b = 10\sqrt{2} \]
Thus, the length of side \( b \) is
\[ \boxed{10\sqrt{2}} \]