Determine the remaining sides and angles of the triangle $A B C$. \[ A=110.47^{\circ}, C=23.27^{\circ}, \mathrm{C}=240 \] \[ B= \] \[ a \approx \] (Do not round until the final answer. Then round to the nearest tenth as needed.) \[ b \approx \] (Do not round until the final answer. Then round to the nearest tenth as needed.)

Step 1 ：We know that the sum of the angles in a triangle is 180 degrees. So we can calculate angle B by subtracting the known angles A and C from 180. \(B = 180 - A - C = 180 - 110.47 - 23.27 = 46.26^{\circ}\)

Step 2 ：We can use the Law of Sines to find the lengths of sides a and b. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

Step 3 ：We can set up the following equations to find a and b: \(\frac{a}{\sin(A)} = \frac{c}{\sin(C)}\) and \(\frac{b}{\sin(B)} = \frac{c}{\sin(C)}\)

Step 4 ：Solving these equations for a and b, we find that \(a = c \cdot \frac{\sin(A)}{\sin(C)} = 240 \cdot \frac{\sin(110.47)}{\sin(23.27)} = 569.1\) and \(b = c \cdot \frac{\sin(B)}{\sin(C)} = 240 \cdot \frac{\sin(46.26)}{\sin(23.27)} = 438.9\)

Step 5 ：Final Answer: The remaining sides and angles of the triangle $ABC$ are: \(B = \boxed{46.26^{\circ}}\), \(a = \boxed{569.1}\), and \(b = \boxed{438.9}\)