Using the Law of Sines to solve the triangle if $\angle A=37^{\circ}, \angle C=69^{\circ}, b=23$ : $\angle B$ is degrees; \[ \begin{array}{l} a=\square ; \\ c=\square ; \end{array} \] Round to two decimal places if needed. Assume $\angle A$ is opposite side $a, \angle B$ is opposite side $b$, and $\angle C$ is opposite side $c$.

Step 1 ：\(\angle B = 180^\circ - \angle A - \angle C = 180^\circ - 37^\circ - 69^\circ = 74^\circ\)

Step 2 ：\(a = b \cdot \frac{\sin A}{\sin B} = 23 \cdot \frac{\sin 37^\circ}{\sin 74^\circ} \approx 15.04\)

Step 3 ：\(c = b \cdot \frac{\sin C}{\sin B} = 23 \cdot \frac{\sin 69^\circ}{\sin 74^\circ} \approx 21.88\)

Step 4 ：Check the solutions: \(\frac{a}{\sin A} = \frac{15.04}{\sin 37^\circ} \approx 23 = b = \frac{23}{\sin 74^\circ}\)

Step 5 ：Check the solutions: \(\frac{c}{\sin C} = \frac{21.88}{\sin 69^\circ} \approx 23 = b = \frac{23}{\sin 74^\circ}\)

Step 6 ：Check the sum of the angles: \(\angle A + \angle B + \angle C = 37^\circ + 74^\circ + 69^\circ = 180^\circ\)

Step 7 ：\(\boxed{\angle B = 74^\circ, a \approx 15.04, c \approx 21.88}\)