Hello wonderful mathematics people. I'm Anna Cox from Kella Community College. Law of cosines is used in oblique triangles to find the angles if we know all three sides or to find the unknown angles inside If we know two sides and an angle but neither of the sides are opposite the known angle. We're going to do this by using coordinate system. We're going to put one of the vertices of the triangle at the origin. If we dropped an altitude along the vertice CB, we know that we've gone over some X amount and up some Y amount. So if we looked at this and we know that this third side over here is C because it's opposite angle C and this is A because it's opposite angle A. And this whole thing down here is because it's opposite angle B. If we look at this right triangle here, however, we know that cosine of angle A would be the adjacent side X over the hypotenuse of C, So X would equal C cosine A. We also know that sine of A would equal the opposite over the hypotenuse, so C sine A would equal the Y. So the coordinate at point B is going to be the X, Y or C cosine A comma C sine A. The coordinate of point A is the origin because we put a vertice of our triangle located there. And the coordinate at Point C is going to be B, 0 because we went over B and we didn't go up and down any. Now what we're going to do is we're going to find the distance A. So the distance formula says A is going to equal the square root of the X coordinates subtracted quantity squared plus the Y coordinates subtracted quantity squared. We're going to square each side, so we're going to have a squared equaling. If we foil this out, we're going to have C ^2 cosine squared A -, 2 BC, cosine a + b ^2 plus the C ^2 sine squared A. Now what we're going to do is some regrouping. If we have this, we're going to regroup. We're going to pull the ones with the C ^2 together, so the C ^2, and I'm going to factor out the C ^2 from the cosine squared A+, the sine squared a, and then I'm going to have the plus b ^2 in there and that -2 BC cosine A. Well, we know that cosine squared A+ sine squared A is really one, so we're going to have a ^2 equaling. If I rewrite it in alphabet 1/4, I'd have b ^2 + C ^2 -, 2 BC cosine A. Now variables are just variables, so they're interchangeable. So we actually have 3 forms of this particular formula. If we have a squared equaling b ^2 + C ^2 -, 2 BC cosine A, we could just redo that triangle, changing the locations of the vertices. So we could also have b ^2 equaling a ^2 + C ^2 -, 2 AC cosine B. Or our third one could be C ^2 equal a ^2 plus b ^2 -, 2 AB, cosine C Now if our triangle was a right triangle, what's cosine of 90°, cosine of 90° is 0. So 0 times this 2 variables at the end, 2 times AC or two times BC or two times AB would go to zero. And then we would actually have our Pythagorean theorem C ^2 equal a ^2 + b ^2 or b ^2 equal a ^2 + C ^2 depending on which one was our hypotenuse or which one was opposite the 90° angle. So if the in this first one, our A was 90°, our side A would be opposite the 90°, and so we'd have a squared equal b ^2 + C ^2. Now taking these formulas, we're going to manipulate them around to solve for an angle also. So those 3 are very important and it doesn't matter where the variable goes as long as the pieces line up. IE if we if we're solving for little A, we have to know the angle A and the other two sides. For solving for little B, we have to know angle B and the other two sides. For solving for side C, we have to know angle C and the other two sides. If I take any of these formulas, I think I'll just start with the top one. If we have a ^2 equaling b ^2 + C ^2 -, 2 BC cosine A, let's say we want to figure out what that angle A equals. I'm going to start by bringing the two BC cosine A over to the left side because I personally like my term to be positive. Then I'm going to subtract the a ^2 to the right side, so b ^2 + C ^2 -, a ^2. I may have said the left and right wrong, but that's what I wanted to do. To get the angle by itself, we're going to divide each side by B2 BC, so cosine A = b ^2 + C ^2 minus a ^2 / 2 BC. And finally, to get the angle by itself, we're going to just take cosine inverse. So A is going to equal cosine inverse of b ^2 + C ^2 -, a ^2 over 2 BC. So cosine inverse is going to be AB squared minus C ^2 oh b ^2 + C ^2. Sorry, pens not working very well. B ^2 + C ^2 -, a ^2 / 2 BC. Now, because variables are interchangeable, we actually would have three of these awesome. We could have angle B equaling cosine inverse of a ^2 + C ^2 -, b ^2 / 2 AC and we could have C equaling cosine inverse of a ^2 + b ^2 -, C ^2 / 2 A2 AB. So basically, if it's the angle over here, we're adding the two sides that are not opposite the angle and subtracting the side that is opposite the angle and dividing by two times the two sides that weren't opposite the angle. Let's look at a couple examples. If we're given A equal 3B equal 4 and C equal 40, so A equal 3, B equal 4, and C was 40° so we're going to find little C We can't use Pythagorean theorem because it's not a right triangle. But we can use law of cosines. So law of cosines says C ^2 equal a ^2 + b ^2 -2 AB cosine C So now if we grab our calculators making sure we're in degree mode, we get 9 + 16 -, 24 cosine 40. So then we're going to square root that and when we square root that, we're going to come up with 2.57. Now at this point, we could actually use law of sines. But if we use law of sines, we have to remember we have to look for two triangles every time. So using law of cosines, we don't have to worry about that. The angle will actually come out whether it's obtuse or acute, because in our unit circle, cosine's only positive from zero to 90 and it's negative from 90 to 180, so it'll come out automatically. So let's find angle A next. If we have A, we can have a equaling cosine inverse of b ^2 plus C ^2. Now, if you want to use an exact, you'd actually type all of this in. So we could say 3 ^2 + 4 ^2 - 2 * 3 * 4 cosine 40, and your calculator will store that for you. And then we're going to subtract the a ^2, so 3 ^2, and that's going to be all over 2B and C I'm just going to write the letter C because I'm going to have it stored in my calculator. So now when we find this, we're going to do cosine inverse. Make sure to put some parentheses in here. So 4 ^2 16 plus, I'm going to go in my calculator and grab that number we had a minute ago minus the 9:00. I'm going to) I'm going to divide. I'm going to use a new set of parentheses to do my 2 * 4 times that C value. And when we hit enter, we should get A being approximately 62.28, so 62.28°. Now we know the three angles add up to 180, but since we're trying to practice a new concept, let's go ahead and find our angle B using our law cosines again. So B equal cosine inverse of a ^2 plus the C ^2 minus the b ^2 all over 2 * a times C And actually when I look at what I just did, I made an error up here. I didn't do my C ^2 when I inputted it in the calculator. I did it here, but I actually did not when I inputted it. So let me go back really quickly and square that C ^2. So this angle really should have been 48.57. I'm sure you all caught that error for me as you were doing it with me. So grabbing my calculator to finish up angle B, I'm going to do cosine inverse of ( 9 plus that C ^2 minus 4 ^2 or 16. I'm going to close that parenthesis. I'm going to divide. I'm going to put another ( 2 * 3 times that C value, which once again I have stored in the calculator. So we're going to get B equaling 91.43. If I've done this right, those 3 angles should add up to 180. Checking on my calculator, they do so literally it's knowing my formulas and just plugging it in. Let's look at an example when we know the three sides and none of the angles. So here we know five, we know 8, and we know 9. So finding any of the angles, let's start with angle A. We're going to have cosine inverse of b ^2 + C ^2 -, a ^2 all over to BC. So grabbing the calculator cosine inverse of a ( 8 ^2 + 9 ^2 - 5 ^2 ) / ( 2 * 8 * 9, we're going to get A being 33.56° if we do B. The formula is the same thing, except now we're changing where things are located. So cosine inverse of a ^2 + C ^2 -, b ^2 all over to AC. Grabbing our calculators, doing cosine inverse ( 5 ^2 + 9 ^2 - 8 ^2 / ( 2 * 5 * 9, we're going to get B being approximately 62.18. And if we do angle CC is going to equal cosine inverse of a ^2 + b ^2 -, C ^2 all over to AB, grabbing our calculators again, cosine inverse ( 5 ^2 + 8 ^2 - 9 ^2 ) divide parentheses divided by 2 * 5 * 80 or 2 * 5 * 8 which is 80. We get C equaling 84.26, so if I've done that correctly, I should be able to take those 3 angles 84.26 + 33.56 + 62.18 and get 180°. Now, may not be exactly 180 because of rounding errors, but it ought to be very, very close. So let's do one more example. Let's go out here and let's look at 1:00. That's got 3-4 and six as its sides. So let's let A equal 3B equal 4C equals 6. If we do 3-4 and six, we're going to have angle a equaling cosine inverse of b ^2 + C ^2 -, a ^2 all over 2 BC. We're going to have B equaling cosine inverse of a ^2 + C ^2 -, b ^2. All over to AC and we're going to have C equaling cosine inverse a ^2 + b ^2 -, C ^2 all over to AB. Now if we grab our calculators and input this so grabbing our calculators we get cosine inverse ( 16 + 36 - 9 / 48 26.38°. If we do the next one cosine inverse ( 9 + 36 - 16 divided by 36 36.34 and if we do the last one cosine inverse ( 9 + 16 - 36 divided by 24 117.28. That was what I wanted to point out was that we really do get an obtuse angle just by doing the law of cosines directly, unlike the law of signs where I have to be testing the angles to make sure I have them. So if we take these three numbers and add them together, so 26.38 plus 36.34 + 117.28, we should get a 180. Thank you and have a wonderful day. This is Annika.