Airports A and B are $492 \mathrm{~km}$ apart, on an east-west line. Jim flies in a northeast direction from A to airport C. From $\mathrm{Che}$ flies $347 \mathrm{~km}$ on a bearing of $126^{\circ} 50^{\prime}$ to $B$. How far is $C$ from $A$ ? The distance between $\mathrm{C}$ and $\mathrm{A}$ is $\square \mathrm{km}$. (Round to the nearest kilometer as needed.)

Step 1 ：Given that airports A and B are 492 km apart, on an east-west line. Jim flies in a northeast direction from A to airport C. From C, he flies 347 km on a bearing of 126°50′ to B.

Step 2 ：We can solve this problem using the law of cosines. The law of cosines states that for any triangle, the square of the length of one side is equal to the sum of the squares of the lengths of the other two sides minus twice the product of the two lengths and the cosine of the included angle.

Step 3 ：In this case, we know the lengths of two sides (492 km and 347 km) and the included angle (126°50′).

Step 4 ：First, we convert the angle to radians: \(2.213659267946141\) radians.

Step 5 ：Then, we use the law of cosines to find the length of the third side (the distance from A to C): \(\sqrt{{492^2 + 347^2 - 2*492*347*\cos(2.213659267946141)}}\)

Step 6 ：After calculating, we find that the distance from A to C is approximately 753 km.

Step 7 ：Final Answer: The distance between C and A is \(\boxed{753}\) km.