It's rather complicated.

d = the distance travelled during 85 degree section.
The distance that the plane traveled initially in the North (positive y) direction is:
d sin (85)
The distance that the plane traveled initially in the East (positive x) direction is:
d cos (85)

When the plane turned in the 200 degree direction, it is now headed Southwest.
The distance the plane travelled in the South direction is:
150 sin (200) (Note : answer will be negative because plane is headed south - negative y direction)
The distance the plane travelled in the West direction is:
150 cos (200) (Note: answer will be negative because plane is headed west - negative x direction)

The final position can be calculated using the Pathagorean Theorem since you know the East/West direction and the North/South direction. Simply add the sides and plug it in.
East/West = (cos 85)d + (150*cos 200)
North/South = (sin 85)d + (150*sin 200)

[(cos 85)d + (150*cos 200)]² + [(cos 85)d + (150*cos 200)]² = 195²

[(cos² 85)d² + 2(cos 85)(150*cos 200)d + 150²*cos² 200] + [(sin² 85)d² + 2(sin 85)(150*sin 200)d + 150²*sin² 200] = 195²

(cos² 85 + sin² 85)d² + [2(cos 85)(150*cos 200)d + 2(sin 85)(150*sin 200)]+ 150²*(cos² 200 + sin² 200) = 195²

Note: cos² x + sin² x = 1

d² + [2(cos 85)(150*cos 200) + 2(sin 85)(150*sin 200)]d + 150² = 195²
d² + [2(cos 85)(150*cos 200) + 2(sin 85)(150*sin 200)]d + 150² = 195²
d² + (-24.570 + -101.216)d + 150² = 195²
d² - 125.786d +22500 = 38025
d² - 125.786d +22500 = 38025
d² - 125.786d – 15525 = 0

Solve using quadratic formula:
d = [-(-125.786) +/- sqrt(125.786² - 4*(-15525)]/2
d = (125.786 +/- 279.145)/2
d = 202.466 mi or -76.680 mi

The answer is 202.466 mi

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