Determine the remaining sides and angles of the triangle $A B C$.
\[
A=28.05^{\circ} \quad B=22.13^{\circ} \quad c=15.14 m
\]
\[
c=
\]

Step 1 ：Given that $A=28.05^{\circ}$, $B=22.13^{\circ}$, and $c=15.14 m$, we can first find the third angle $C$ using the fact that the sum of angles in a triangle is $180^{\circ}$. So, $C = 180^{\circ} - A - B = 180^{\circ} - 28.05^{\circ} - 22.13^{\circ} = 129.82^{\circ}$.

Step 2 ：Next, we can use the Law of Sines to find the other two sides. 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. So, we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.

Step 3 ：We can use this to find $a$ and $b$. For $a$, we have $a = \frac{c \sin A}{\sin C} = \frac{15.14 m \sin 28.05^{\circ}}{\sin 129.82^{\circ}} = 7.96 m$.

Step 4 ：For $b$, we have $b = \frac{c \sin B}{\sin C} = \frac{15.14 m \sin 22.13^{\circ}}{\sin 129.82^{\circ}} = 6.27 m$.

Step 5 ：So, the remaining sides and angles of the triangle $ABC$ are $a = 7.96 m$, $b = 6.27 m$, and $C = 129.82^{\circ}$.