Solve the triangle. \[ B=72^{\circ} 51^{\prime}, c=31 m, a=80 m \] What is the length of side $b$ ? \[ \mathrm{b}=\square \mathrm{m} \] (Round to the nearest whole number as needed.) What is the measure of angle $A$ ? \[ A= \] (Round to the nearest whole number as needed.) What is the measure of angle $\mathrm{C}$ ? \[ \mathrm{C}= \] (Round to the nearest whole number as needed.)

Step 1 ：We are given two sides and an included angle of a triangle. We can use the Law of Cosines to find the length of side $b$. The Law of Cosines states that for any triangle with sides of lengths $a$, $b$, and $c$ and an angle $B$ opposite side $b$, the following relationship holds: $b^2 = a^2 + c^2 - 2ac \cos B$. We can rearrange this formula to solve for $b$: $b = \sqrt{a^2 + c^2 - 2ac \cos B}$. We can substitute the given values into this formula to find the length of side $b$. After calculation, we find that $b = \boxed{77}$ m.

Step 2 ：Now, we can use the Law of Sines to find the measures of angles $A$ and $C$. 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. This can be written as: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. We can rearrange this formula to solve for $A$ and $C$: $A = \sin^{-1}\left(\frac{a \sin B}{b}\right)$ and $C = \sin^{-1}\left(\frac{c \sin B}{b}\right)$. We can substitute the given values into these formulas to find the measures of angles $A$ and $C$. After calculation, we find that $A = \boxed{83}$ degrees and $C = \boxed{23}$ degrees.