Source: 1976 Euclid Part B Problem 1
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Triangle $ABC$ has $\angle{B}=30^{\circ}$, $AB=150$, and $AC=50\sqrt{3}$. Determine the length of $BC$.
1. Given the triangle \(ABC\) with \(\angle B = 30^\circ\), \(AB = 150\), and \(AC = 50\sqrt{3}\), we need to determine the length of \(BC\).
2. First, let's verify if \(\angle A = 90^\circ\) to see if it forms a \(30^\circ-60^\circ-90^\circ\) triangle. In a \(30^\circ-60^\circ-90^\circ\) triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\).
3. If \(\angle A = 90^\circ\), then \(\angle C = 60^\circ\) because the sum of angles in a triangle is \(180^\circ\).
4. In a \(30^\circ-60^\circ-90^\circ\) triangle, the side opposite the \(30^\circ\) angle is half the hypotenuse. Here, \(AB\) would be the hypotenuse, and \(AC\) would be the side opposite the \(60^\circ\) angle.
5. Let's check the given lengths:
   - Hypotenuse \(AB = 150\)
   - Side opposite \(60^\circ\) (AC) should be \( \frac{150\sqrt{3}}{2} = 75\sqrt{3} \)
6. However, \(AC\) is given as \(50\sqrt{3}\), which does not match \(75\sqrt{3}\). Therefore, \(\angle A\) is not \(90^\circ\), and the triangle is not a \(30^\circ-60^\circ-90^\circ\) triangle.
7. Next, consider the possibility of a \(30^\circ-30^\circ-120^\circ\) triangle. In this case, \(\angle A = 30^\circ\) and \(\angle C = 120^\circ\).
8. Using the Law of Cosines to find \(BC\):
   \[
   BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)
   \]
   \[
   BC^2 = 150^2 + (50\sqrt{3})^2 - 2 \cdot 150 \cdot 50\sqrt{3} \cdot \cos(30^\circ)
   \]
   \[
   BC^2 = 22500 + 7500 - 15000\sqrt{3} \cdot \frac{\sqrt{3}}{2}
   \]
   \[
   BC^2 = 22500 + 7500 - 15000 \cdot \frac{3}{2}
   \]
   \[
   BC^2 = 22500 + 7500 - 22500
   \]
   \[
   BC^2 = 7500
   \]
   \[
   BC = \sqrt{7500} = 50\sqrt{3}
   \]

The final answer is \( \boxed{ 50\sqrt{3} } \)