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Problem: A dog in an open field runs 12.0m east and then 30.0m in a direction 54 degrees west of north.

Part A: In what direction must the dog then run to end up 12.0m south of her original starting point?

My Answer: Let the first path the dog runs A, the second path B, the third path C, and $R=A+B+C$.

$A_x=12$, $A_y=0$, $B_x=-30\cos(54)$, $B_y=30\sin(54)$, $R_x=0$, and $R_y=-12$

So $C_x=R_x-A_x-B_x=0+30\cos(54)-12$ and $C_y=R_y-A_y-B_y=-12-30\sin(54) + 0$.

Thus, $\arctan(30\cos(54)-12)/(-12-30\sin(54))$ which is equal to -8.8 east of south.

Is that correct?

Part B: How far must the dog then run to end up 12.0 south of her original starting point?

Using the components above, we find the magnitude of C.

The magnitude of C is $\sqrt{(30\cos(54)-12)^2+(-12-30\sin(54))^2}$ which is equal to 30.2m

Is that correct?

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How close to the South Pole is this field located? – DJohnM Aug 30 '13 at 5:33
up vote 1 down vote accepted

Almost. Note that the angle of path $B$ is $54^\circ$ West of North (not North of West). Hence, you mixed up your $\sin$ and $\cos$. You should obtain: \begin{align*} C_x &= 30\sin(54^\circ)-12 \\ C_y &= -30\cos(54^\circ)-12 \\ \theta &= \arctan(|C_x/C_y|) = \arctan \left(\frac{30\sin(54^\circ)-12}{30\cos(54^\circ)+12} \right) =22.49...^\circ \text{ East of South} \\ C &= \sqrt{C_x^2 + C_y^2} = \sqrt{(30\sin(54^\circ)-12)^2 + (-30\cos(54^\circ)-12)^2} = 32.07... \text{metres} \end{align*}

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Thank you for the help. I feel so silly about it – Username Unknown Aug 30 '13 at 0:54
Thanks for the edit. And you're welcome. =] – Adriano Aug 30 '13 at 1:08

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