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An air traffic controller observes two airplanes approaching the airport. The displacement from the control tower to plane 1 is given by the vector A⃗ , which has a magnitude of 220 km and points in a direction 32 ∘ north of west. The displacement from the control tower to plane 2 is given by the vector B⃗ , which has a magnitude of 140 km and points 65 ∘ east of north.

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HomertheGenius
We will use the Cosine Law:
m D² = A² + B² - 2 AB cos D
∠D = 65° + 58° = 123° 
m D² = 220² + 140² - 2 * 220 * 140 * cos 123°
m D² = 48,000 + 19,600 + 33,854.72
m D = 319.15 km
After that we will use the Sine Law:
340 / sin 123° = 220 / sin (theta)
sin (theta) = 0.578
∠(theta) = sin^(-1)0.578 = 35.3°
Answer: 
The magnitude of the vector D is 319.15 km and points 35.3° south to east.