I dont get it help me please

Question

Find the magnitude and direction of the resultant vector that describes the overall motion of a plane that travels 340 km at 12° East of North, then turns to travel 40 km at 34° South of East

Answer 1

The statement of the problem may be a little confusing, so let me restate my understanding

of it in terms of the x-y coordinates.

As usual, the positive x direction represents due East and the positive y direction

represents due North.

The plane starts from the origin point with (x, y) coordinates (0,0).

It travels 340 km at 12° to the right off the positive vertical direction (y-axis).

That is 78° off the standard positive x-axis.

That leg of the trip forms the vector (340 cos(78°), 340 sin(78°)).

From there it travels 40 km at 34° below the positive horizontal direction,

that is -34° off the standard positive x-axis.

That leg of the trip forms the vector (40 cos(-34°), 40 sin(-34°)).

The resultant vector is the sum

(340 cos(78°) + 40 cos(-34°), 340 sin(78°) + 40 sin(-34°)) = (103.85, 310.2)

The magnitude of the vector is

√(103.85² + 310.2²) = 327.112

The direction of the resultant vector is the angle it forms off the positive x-direction:

arctan(310.2 / 103.85) = 71.49°.