Let us take northern direction as positive y, and eastern direction as positive x. Then the position of the first plane is given by the coordinates:
The northern component of the velocity of the first plane is 400*cos(60), the eastern one is 400*sin(6).
For the second plane, the eastern component of the velocity is given by 425*sin(40), northern component is given by -425*cos(40). Minus sign indicates that it is directed southward, as it should be according to the problem's statement. Thus, coordinates of the second plane are:
In three hours, the first plane will be in the point A with coordinates (1039.23; 600), the second plane will be at the point B (819.55; -976.71). The distance between those points can be found by the formula:
If two planes fly towards each other, then the speed of approach is the sum of two speeds, that is 825 mph. Divide the distance D by 825 to obtain the time until collision, tc=1592/825≈1.93 hours or 1 hour 55 minutes and 47 seconds. This answers the first question.
The answer to the second question can be obtained by multiplying the time found in the first question, tc, by the velocity of each plane. So we obtain that the first plane will travel 400*1.93=772 miles, the second plane will travel 425*1.93=820 miles. These are the distances traveled from the point of turn to the collision point.