We start by finding angles within the triangle:
- first, the helicopter switches from a heading of 34° to one of 162°
While 34° is a 1st quadrant angle, as the helicopter exits that heading to embark on a new one we can envision it as coming in toward the origin from the 3rd quadrant, still making a 34° angle coming down from the x-axis into that 3rd quadrant. 162° is a 2nd quadrant angle, with a supplementary angle of 180°-162° = 18° coming up into the 2nd quadrant.
Bringing this all together, we find we have a 34°+18° = 52° angle as the first angle of our triangle; let's call this angle A.
- at its next change, the helicopter switches from the 162° heading to one of 308°
Still recognizing 162° as making an 18° angle with the x-axis, we're now exiting this heading as if it's in the 4th quadrant, coming 18° down from the x-axis. 308° is also in the 4th quadrant, making a 360°-308° = 52° angle with the x-axis.
Since we are cutting the 18° angle out of the same quadrant instead of tacking it on through a further quadrant like in the situation above, this time we subtract to find a 52°-18° = 34° angle as the second angle in our triangle; call it B.
Notice that we also know the side length b opposite angle B: it is given as 50 miles. We know one side and two angles, and in particular since what we want is the length of the side opposite angle A, we are ready to use the Law of Sines to finish the problem!
sin(A)/a = sin(B)/b
so
sin(52)/a = sin(34)/50
take the reciprocal of both sides, then multiply both sides by sin(52) to find
a = sin(52)*50/sin(34) ≈ 70.46 miles
