An airplane flies 20km in a direction 60 degrees north of east, then 30 km straight east, then 10km straight north. how far and in what direction is the plane from the starting point

48.4 km, 34.3° north of east

Explanation:

Let's say east is the +x direction and north is the +y direction.

Adding up the x components of the vectors:

x = 20 cos 60 + 30 + 0

x = 40 km

Adding up the y components of the vectors:

y = 20 sin 60 + 0 + 10

y = 27.3 km

The magnitude of the displacement is:

d = √(x² + y²)

d = 48.4 km

The direction is:

θ = atan(y/x)

θ = 34.3° north of east

i cant see the choices man

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