This first step involves a right triangle. If the plane is 400 km east and 300 km south of the origin, and it flew in a straight line, then you can construct a right triangle with side lengths 300, 400, and c. You may recognize that these are multiples of the Pythagorean triple 3, 4, 5, so the side length c is 500 km. Otherwise, you would write
[tex]c^2 = 300^2+400^2 \\ c = 500[/tex].
This second step is, if I am correctly interpreting "degrees south of east," to find the angle formed by the horizontal line representing the east and the path of the plane. I made a diagram that does just that (see attached). You can use a trig function of one of the angles to solve. I chose
[tex]tan(C)= \frac{3}{4} \\ arctan( \frac{3}{4})=36.870[/tex]. Thus, I believe it is 37° south of east.