phys questions referring vectors

Question

Three horizontal ropes pull on a large stone stuck in the ground, producing the vector forces A⃗ , B⃗ , and C⃗ shown in the figure below

Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero. Part A I got 90.2 N
from:

Rx = 100cos30° + 80cos120° + 40cos233° + Dx = 0

Dx = -22.53N

Ry = 100sin30° + 80sin120° + 40sin233° + Dy = 0

Dy = -87.34N

then using magnitude function(sqrt(a^2+b^2) = 90.2N

Find the direction of a fourth force on the stone that will make the vector sum of the four forces zero. For Part B I used arctan[(-87.34N)/(-22.53N)] = 75.5° but it is wrong. Why?

Answer 1

Assuming your computation of the fourth force being $(-87.34,-22.54)$ is correct, then what you want is $75.5^\circ+180^\circ$. It may help to note that $\arctan(-87.34/(-22.53))=\arctan(87.34/22.53)$, which will give you one of the angles of a right triangle with legs $87.34$ and $22.53$. However, your fourth force lies in the 3rd quadrant, so you should add $180^\circ$.

Answer 2

To avoid mistakes about the directions of vectors, if you have $x,y$ coordinates for a vector (as you do in the case of your vector $D$), use those to plot the vector first.

Since you found $D_x = -22.53$ N and $D_y = -87.34$ N, try plotting the point $(-22.53, -87.34)$ on a diagram of the forces. You should find that it points down and to the left, but much more down than left.

Now that you have an approximate idea of the direction of the vector, you can recognize whether the calculated value is reasonable. Remember that every trigonometric function takes most of its values twice as you go from $0$ to $360^\circ$, so you always have to be alert to whether you're getting the angle you want and not the "other" angle.