A question on my trigonometric test a student solved using both law of sines and law of cosines and got similar but different results (I've double checked the calculations and cannot think of a reason they would be different). Vector one $~12~$ lb pulls on a sled at an angle of $~15^{\circ}~$ "above" the $~x$-axis, vector two $~18~$ lbs pulls on the the sled at an angle of $~10^{\circ}~$ "below" the $~x$-axis. What is the magnitude of the resultant force?
The law of cosines calculation where $x$ is the magnitude of the diagonal of the parallelogram yields $x^2 = 12^2+18^2-2(12)(18) \cos(155)$
$x = 29.3176$ lb
If using the law of sines:
$\frac{\sin(155)}{x} = \frac{\sin(10)}{12}$
$x = 29.2051$ lb
If the $~15^{\circ}~$ angle is used instead of the $~10^{\circ}~$ the answer is a little higher than law of cosines answer. Breaking the vectors up into $~x~$ and $~y~$ components, adding them and finding the magnitude yields the same answer as law of cosines.
Best Answer
It was a good idea to do the problem a third way -- it gives more credence to the law of cosines answer than to the law of sines answer. Additionally, you seem to be somewhat uncomfortable with the law of sines method -- you mention that you consider using $25$ degrees instead of $10$ degrees, so you are at least somewhat unsure what that angle should be.
You're thus very close to the root of the issue: the angle between the parallelogram's diagonal and the bottom vector is unknown. It's not $10$ degrees (that's the angle between the x-axis and the bottom vector), and it's not $25$ degrees (that's the angle between the top vector and the bottom vector). We can't apply the law of sines because we don't have two known angles.