### Instruction: In triangle $ABC,$ $\angle B = 30^\circ,$ $AB = 150,$ and $AC = 50 \sqrt{3}.$  Find the sum of all possible values of $BC.$.### Response: This problem involves some trigonometry and the law of cosines.
The law of cosines relates the lengths of the sides of a triangle and the cosine of the angle opposite one of them.
In this case, I can use the law of cosines to find $BC$ by plugging in the given values of $AB,$ $AC,$ and $\angle B.$
The law of cosines says that $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos B.$
Substituting the given values, I get $BC^2 = 150^2 + (50 \sqrt{3})^2 - 2 \cdot 150 \cdot 50 \sqrt{3} \cdot \cos 30^\circ.$
Simplifying, I get $BC^2 = 22500 + 7500 - 15000 \sqrt{3} \cdot \frac{\sqrt{3}}{2}.$
Further simplifying, I get $BC^2 = 30000 - 22500 = 7500.$
Taking the square root of both sides, I get $BC = \pm \sqrt{7500}.$
However, $BC$ has to be a positive length, so I only consider the positive root.
Therefore, $BC = \sqrt{7500}.$