A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 254 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle è with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle è.

I know that for the magnitude, since both vectors have different directions I cannot just add them together. I am not sure where to go from here.

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1. Divide both vectors into East (x) and North (y) components

First vector:
east1 = 254 cos 30 = 220
north1 =254 sin 30 = 127

Second vector
east 2 = - 178
north2 = 0

now add east components
220 - 178 = 42 east
now the north components
127 + 0 = 127
so
magnitude = sqrt (42^2+127^2) = 134
tan e = 127/42 = 3.02
so e = 71.7 degrees north of east

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