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

Step 1 ：Given that $A={28.05}^{\circ }$, $B={22.13}^{\circ }$, and $c=15.14m$, 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 $a=\frac{c\sin A}{\sin C}=\frac{15.14m\sin {28.05}^{\circ }}{\sin {129.82}^{\circ }}=7.96m$.

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

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