Writing $1989$ in base $b$, we obtain a three-digit number: $xyz$. It is known that the sum of the digits is the same in base $10$ and in base $b$, that is, $1 + 9 + 8 + 9 = x + y + z$. Determine $x,y,z,b.$ 
1. We start by noting that the number \(1989_{10}\) is written as a three-digit number \(xyz\) in base \(b\). This implies:
   \[
   b^2 \leq 1989 < b^3
   \]
   Solving for \(b\), we get:
   \[
   b^2 \leq 1989 \implies b \leq \sqrt{1989} \approx 44.6 \implies b \leq 44
   \]
   \[
   1989 < b^3 \implies b > \sqrt[3]{1989} \approx 12.6 \implies b \geq 13
   \]
   Therefore, the range for \(b\) is:
   \[
   13 \leq b \leq 44
   \]

2. Given that the sum of the digits in base \(10\) and base \(b\) are the same, we have:
   \[
   1 + 9 + 8 + 9 = x + y + z \implies 27 = x + y + z
   \]

3. The number \(1989_{10}\) can be expressed in base \(b\) as:
   \[
   x \cdot b^2 + y \cdot b + z = 1989
   \]

4. Substituting \(x + y + z = 27\) into the equation, we get:
   \[
   x \cdot b^2 + y \cdot b + z = 1989
   \]
   \[
   x \cdot b^2 + y \cdot b + (27 - x - y) = 1989
   \]
   Simplifying, we get:
   \[
   x \cdot b^2 + y \cdot b + 27 - x - y = 1989
   \]
   \[
   x \cdot (b^2 - 1) + y \cdot (b - 1) = 1962
   \]

5. Factoring out \((b-1)\), we get:
   \[
   (b-1)[(b+1)x + y] = 1962
   \]
   Since \(1962 = 18 \times 109\), we need to find \(b\) such that \((b-1)\) is a factor of \(1962\). Testing possible values within the range \(13 \leq b \leq 44\), we find:
   \[
   b = 19
   \]

6. Substituting \(b = 19\) into the equation, we get:
   \[
   (19-1)[(19+1)x + y] = 1962
   \]
   \[
   18[20x + y] = 1962
   \]
   \[
   20x + y = \frac{1962}{18} = 109
   \]

7. Solving for \(x\) and \(y\) with \(x + y + z = 27\), we find:
   \[
   20x + y = 109
   \]
   \[
   x + y + z = 27
   \]
   Substituting \(z = 27 - x - y\), we get:
   \[
   20x + y = 109
   \]
   \[
   x + y + (27 - x - y) = 27
   \]
   Solving for \(x\) and \(y\), we find:
   \[
   x = 5, \quad y = 9, \quad z = 13
   \]

8. Verifying, we have:
   \[
   5 \cdot 19^2 + 9 \cdot 19 + 13 = 1989
   \]
   \[
   5 \cdot 361 + 9 \cdot 19 + 13 = 1989
   \]
   \[
   1805 + 171 + 13 = 1989
   \]
   \[
   1989 = 1989
   \]

Thus, the values are \(x = 5\), \(y = 9\), \(z = 13\), and \(b = 19\).

The final answer is \( \boxed{ (5, 9, 13, 19) } \)