Determine the remaining sides and angles of the triangle $ABC$.
$$\mathrm{c}=7\mathrm{mi},\mathrm{B}={38.17}^{\circ },\mathrm{C}={34.28}^{\circ }$$

Step 1 ：Given that side c = 7 mi, angle B = 38.17 degrees, and angle C = 34.28 degrees in triangle ABC.

Step 2 ：First, we can find angle A by subtracting the given angles B and C from 180 degrees. So, $A=180-B-C=180-38.17-34.28=107.55$ degrees.

Step 3 ：Next, 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 4 ：Using the Law of Sines, we can find side a as follows: $a=c\cdot \frac{\sin (A)}{\sin (C)}=7\cdot \frac{\sin (107.55)}{\sin (34.28)}=11.85$ mi.

Step 5 ：Similarly, we can find side b as follows: $b=c\cdot \frac{\sin (B)}{\sin (C)}=7\cdot \frac{\sin (38.17)}{\sin (34.28)}=7.68$ mi.

Step 6 ：$\boxed{{\displaystyle \text{Final Answer: The remaining sides and angles of the triangle ABC are }A={107.55}^{\circ },a=11.85\text{ mi},\text{ and }b=7.68\text{ mi}.}}$