Let $\triangle ABC$ be an equilateral triangle and $P$ a point on $\overline{BC}$.  If $PB=50$ and $PC=30$, compute $PA$.
1. **Using the Law of Cosines:**

   Given that $\triangle ABC$ is an equilateral triangle, each side is equal. Let $AC = AB = BC = 80$ (since $PB = 50$ and $PC = 30$, $BC = PB + PC = 50 + 30 = 80$).

   The Law of Cosines states:
   \[
   PA^2 = AC^2 + PC^2 - 2 \cdot AC \cdot PC \cdot \cos(\angle ACP)
   \]
   Since $\triangle ABC$ is equilateral, $\angle ACP = 60^\circ$. Therefore:
   \[
   \cos(60^\circ) = \frac{1}{2}
   \]
   Substituting the values:
   \[
   PA^2 = 80^2 + 30^2 - 2 \cdot 80 \cdot 30 \cdot \frac{1}{2}
   \]
   Simplifying inside the square root:
   \[
   PA^2 = 6400 + 900 - 2400
   \]
   \[
   PA^2 = 4900
   \]
   Taking the square root of both sides:
   \[
   PA = \sqrt{4900} = 70
   \]

2. **Using Stewart's Theorem:**

   Stewart's Theorem for a triangle states:
   \[
   AB^2 \cdot PC + AC^2 \cdot PB = AP^2 \cdot BC + BP \cdot PC \cdot BC
   \]
   Substituting the given values:
   \[
   80^2 \cdot 30 + 80^2 \cdot 50 = AP^2 \cdot 80 + 50 \cdot 30 \cdot 80
   \]
   Simplifying:
   \[
   6400 \cdot 30 + 6400 \cdot 50 = AP^2 \cdot 80 + 50 \cdot 30 \cdot 80
   \]
   \[
   192000 + 320000 = AP^2 \cdot 80 + 120000
   \]
   \[
   512000 = AP^2 \cdot 80 + 120000
   \]
   Subtracting 120000 from both sides:
   \[
   392000 = AP^2 \cdot 80
   \]
   Dividing both sides by 80:
   \[
   AP^2 = 4900
   \]
   Taking the square root of both sides:
   \[
   AP = \sqrt{4900} = 70
   \]

The final answer is $\boxed{70}$