Given vectors $A= 2km [N]$, $B=3km [W]$ and $C=5 km [W 60 S]$ find $a + b + c$ (resultant)? Given vectors A= 2km [N], B=3km [W] and C=5 km [W 60 S] find a + b + c (resultant)?
Attempt
So I used components to solve for the resultant. I also kept in mind significant digits.
$$Ry = Cy + Ay\\
Ry = (5 \cos 30^\circ) - 2 km [S]\\
Ry = 2 km [S]\\
Rx = Cx + Bx\\
Rx = (5 \sin 30^\circ) + 3 km [W]\\
Rx = 6 km [W]\\$$
Therefore continuing pythagorean theorem
$$R^2 = \sqrt(6km^2 + 2km^2)\\
R = 6 km\\
\tan a = 2/6\\
a = 22^\circ$$
My result was:
$$6 km [W 22 S]$$
Did I get this right? Thank you in advance!
 A: Your method looks right. Essentially you're:

*

*Resolving the vectors onto the vertical (facing S) and horizontal (facing W) axes.


*Then you're working out the magnitude of the resultant vector using Pythagorus.


*Finally you're working out the angle $\theta$ that that vector makes with the horizontal (W) axis, by saying $\tan \theta = R\sin\theta / R\cos\theta = R_y/R_x$.
So the method is good. One initial comment though: it would be great to include a sketch (I did one myself working through your problem).
But the bad news is that you've made several computational errors.

*

*$R_y = 5 \cos 30 - 2 \neq 2$ and $R_x = 5 \sin 30 + 3 \neq 6$. Just run these through your calculator and you should get the right numbers.


*Strictly speaking it's $R^2 = (R_x^2 + R_y^2)$, i.e. you shouldn't introduce the square root until you take the square root of $R^2$ as well, i.e. $R = \sqrt{R_x^2 + R_y^2}$ is ok.
Also, $\sqrt{6^2 + 2^2} \neq 6$. But going back to point 1, the numbers need to be different anyway.


*I'm not sure that $\tan a = 2/6 \implies a = 22$ but I may have this wrong. In any case $\tan a \neq 2/6$ because of the numbers above.
All in all, good job, because with the right method you'd get most of the marks in an exam, but you could get those last few marks by practising using a calculator.
