How many noncongruent triangles are there with one side of length $20,$ one side of length $17,$ and one $60^{\circ}$ angle?
To determine the number of noncongruent triangles with one side of length \(20\), one side of length \(17\), and one \(60^\circ\) angle, we will consider three cases based on the position of the \(60^\circ\) angle.

1. **Case 1: The \(60^\circ\) angle is between the sides of length \(20\) and \(17\).**

   By the Law of Cosines, we have:
   \[
   x^2 = 20^2 + 17^2 - 2 \cdot 20 \cdot 17 \cdot \cos(60^\circ)
   \]
   Since \(\cos(60^\circ) = \frac{1}{2}\), the equation becomes:
   \[
   x^2 = 400 + 289 - 2 \cdot 20 \cdot 17 \cdot \frac{1}{2}
   \]
   \[
   x^2 = 400 + 289 - 340
   \]
   \[
   x^2 = 349
   \]
   \[
   x = \sqrt{349}
   \]
   Therefore, one possible triangle has sides \(20\), \(17\), and \(\sqrt{349}\).

2. **Case 2: The \(60^\circ\) angle is between the sides of length \(20\) and \(x\).**

   By the Law of Cosines, we have:
   \[
   17^2 = 20^2 + x^2 - 2 \cdot 20 \cdot x \cdot \cos(60^\circ)
   \]
   Since \(\cos(60^\circ) = \frac{1}{2}\), the equation becomes:
   \[
   289 = 400 + x^2 - 20x
   \]
   \[
   289 = 400 + x^2 - 10x
   \]
   \[
   x^2 - 10x + 111 = 0
   \]
   Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -10\), and \(c = 111\):
   \[
   x = \frac{10 \pm \sqrt{100 - 444}}{2}
   \]
   \[
   x = \frac{10 \pm \sqrt{-344}}{2}
   \]
   Since the discriminant is negative, there are no real solutions for \(x\) in this case.

3. **Case 3: The \(60^\circ\) angle is between the sides of length \(17\) and \(x\).**

   By the Law of Cosines, we have:
   \[
   20^2 = 17^2 + x^2 - 2 \cdot 17 \cdot x \cdot \cos(60^\circ)
   \]
   Since \(\cos(60^\circ) = \frac{1}{2}\), the equation becomes:
   \[
   400 = 289 + x^2 - 17x
   \]
   \[
   400 = 289 + x^2 - \frac{17x}{2}
   \]
   \[
   x^2 - 17x + 111 = 0
   \]
   Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -17\), and \(c = 111\):
   \[
   x = \frac{17 \pm \sqrt{289 - 444}}{2}
   \]
   \[
   x = \frac{17 \pm \sqrt{-155}}{2}
   \]
   Since the discriminant is negative, there are no real solutions for \(x\) in this case.

Therefore, the only valid triangle is the one from Case 1 with sides \(20\), \(17\), and \(\sqrt{349}\).

The final answer is \(\boxed{1}\)