A plane flying 16 degrees west of north at 289mph is flying in air which is moving 32mph at due west. Find the resulting speed and direction of the plane. (Step by step please!)

Answer 1

I would just add up the two vectors
(289cos106°, 289sin106°) + (32cos180, 32sin180)
= (-79.659 , 277.805) + (-32 , 0)
= (-109.659 , 277.805)

magnitude = √(109.659^2 + 277.805^2)
= 298.665 mph

direction ?
tanØ = 277.805/-109659
Ø = 111.54°

which can be expressed as N 21.5° W

Answer 2

A plane is heading 25° west of south. How many degrees south of west is this plane flying?