Help! Find the Magnitude of a Displacement Vector

[DavidAp]

A woman walks 143m in the direction 55° east of north, then 178m directly east. Find the magnitude of the displacement vector.

Answer: 306m

Relevant equations:
I will use vA as a shorthand to represent vector A and ||vA|| to represent the magnitude.

Ax = ||vA||cos(theta)
Ay = ||vA||sin(theta)
||vA|| = sqrt(Ax^2 + Ay^2)

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Since I know the magnitude of vA (assuming vA is the starting vector) and the angle in which she left the origin of the coordinate grid I can use the two equations stated above to find the values of Ax and Ay.

Ax = 143m*cos(55) = 82.02m
Ay = 143m*sin(55) = 117.14m

Now I add vA + vB (<82.02m, 117.14m> + <178m, 0m>) to obtain vC, the vector displacement between her starting position to her final position. However, when I go and find the magnitude of the vector I always come out with the wrong answer, 285.19m.

vC = <260.02m, 117.14m>
||vC|| = sqrt(260.02m^2 + 117.14m^2) = sqrt(81332.18m^2) = 285.19m.

What did I do wrong? I thought this would be a simple problem but I keep coming out with the wrong answer. Can someone help me?

 

[Zyxer22]

The problem is that you're angle is wrong. Try drawing a small graph with the angle starting from the positive y-axis (north) and going towards the x-axis (east). You see that the 55° angle is made with the y-axis, and not the x-axis.

 

[Jaynte]

Yes the angle is wrong. Since they say east of north (which I personally think is a stupid way to say) they mean 90-55=35 degrees. The rest is correct.