The triangle on the right has measurements of 90degrees (where the line meets the base), 7degrees (that you calculated before, and the top is 180-90-7 = 83degrees.
PHEW!
Now we can start solving the problem. Let's focus on the first triangle on the left. We know the length of one of the sides (it's 80 miles). We also know the measures of all the angles, and it's a right triangle. So lets use a trig function. cos13 = x/80. so x = 80cos13. solve for x, and that will tell you the length of the normal flying course that spans this part of the triangle.
Now for the second triangle, we don't know any lengths. Oh wait, we can solve for one - the line we drew to split the original triangles, is a line that both two triangles share. sin13 = center line/80.
center line = 80sin13. Solve for that, and now you know that distance.
So for the second triangle, tan7 = center line/y. that means y = (center line)/tan7. Solve for y.
the original distance is x + y
Now the distance of the flight taken is 80 + the top left line of the original triangle (lets call that LINE). Lets solve for that line using trig functions. sin7 = (center line)/LINE. That means LINE = (center line)/sin7. remember, we solved for the value of center line earlier. Just plug and chug to solve for LINE.
distance travelled = LINE+ 80.
Hope that helps. These problems aren't hard, just read the problem and do exactly what it tells you. If you get the core concepts and few tricks, things will fall into place.