>>Alright so we have the links of all the sides of the triangle. Now we decide that this is the longest side so we're gonna solve for the angle opposite that side and that would be angle B. Now notice that this is a version of and there are 3 versions of the law of cosines in this form. It could be cosine B right here, cosine A or cosine C and we have to kind of structure the fraction over here accordingly. We have the different sides here and the one that's being subtracted, this minus B squared, the B is opposite that angle. That's kind of the important key point here and the AC, the two sides in the denominator here come from these 2 rascals in the numerator. Ok so that's kind of how you'll remember it. Anyway you just plug in the information. We find cosine B to be this fraction and now we want the angle whose cosine is that so we want inverse cosine of that fraction and we find it to be approximately 116.8 degrees, so it's obtuse. So that means then that, oh and by the way I almost forgot to mention, there's only 1 angle here when we're using this law of cosines. When we come to an answer right here we don't have this situation where we have the possibility of a first or second quadrant angle whose cosine is the same value and stuff like that, like we do with the law of sines and that has to do with, well it has to do with a lot of discussions that we had before but remember that the domain for our values here, our angles, when we talk about cosine, fall into first and second quadrant you see. And so we have kind of built in the ideal that we can have kind of acute or obtuse angles that emerge from the calculation. Ok so we don't have that ambiguous possibility here when we use this. Alright so we have an obtuse angle here. One of the angles is obtuse. The other 2 must be acute and so we can go through and use the law of sines without the ambiguous possibility. Ok now let's find the other angles. Let's find the angle opposite, oh let's say side A. So it would be angle A so we write, sine A over A equals sine B over B, we just found the size of that angle. So B is 116.8, side B is 19, side A is 8 and then multiplying by 8 on both sides we get this fraction. So do we want the angle whose sine is that fraction and we find that angle to be approximately 22.1 degrees. Now notice I'm using a little more accuracy here in this problem. Really you can use whatever amount of accuracy that you want to in problems like this. You can carry this out to 3 or 4 decimal places. Just be consistent from problem to problem when you do that. Also I would suggest that when you plug in the value here for this angle that you use a few more digits here to describe this angle. That will give you a little more accuracy as well. Ok at any rate we find angle A to be 22.1 degrees and we need not worry about a second quadrant reference angle of 22.1 degrees and all of that business. We know that we have an acute angle here within our triangle. And then for angle C it's 180 minus the other 2 so it's 180 minus 116.8 minus 22.1 or approximately 41.1 degrees.