Determine the remaining sides and angles of the triangle $A B C$.
\[
A=140^{\circ} 50^{\prime}, C=10^{\circ} 10^{\prime}, A B=10
\]
\[
B=\square
\]
$B C \approx$
(Do not round until the final answer. Then round to the nearest hundredth as needed.)
$A C \approx$
(Do not round until the final answer. Then round to the nearest hundredth as needed.)

Step 1 ：Given that the angles A and C of the triangle are \(140^{\circ} 50^{\prime}\) and \(10^{\circ} 10^{\prime}\) respectively, and the side AB is 10.

Step 2 ：We know that the sum of the angles in a triangle is \(180^{\circ}\). So we can calculate angle B by subtracting the given angles A and C from \(180^{\circ}\).

Step 3 ：Using the law of sines, which states that the ratio of the length of a side of a triangle to the sine of its opposite angle is the same for all three sides of the triangle, we can set up the following equations to find the lengths of the sides BC and AC: \(\frac{BC}{\sin(A)} = \frac{AB}{\sin(C)}\) and \(\frac{AC}{\sin(B)} = \frac{AB}{\sin(C)}\).

Step 4 ：Solving these equations, we find that BC is approximately 35.78 and AC is approximately 27.47.

Step 5 ：Thus, the remaining angle and sides of the triangle are: \(B = \boxed{29.0^{\circ}}\), \(BC \approx \boxed{35.78}\), and \(AC \approx \boxed{27.47}\).