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# Solving for a side with the law of sines

CCSS.Math:

## Video transcript

we've got a triangle here where we know two of the angles and one of the sides and what I claim is that I can figure out everything else about this triangle just with this information you give me two angles and a side and I can figure out what the other two sides are going to be and I can of course figure out the third angle so let's let's try to figure that out and the way that we're going to do it we're going to use something called the law of sines and in a future video I will prove the law of sines but here I am just going to show you how we can actually apply it and it's a fairly straightforward idea the law of sines just tells us that the ratio between açaí the sine of an angle and the side opposite to it is going to be constant for any of the angles in a triangle so for example for this triangle right over here this is a 30-degree angle this is a 45 degree angle they have to add up to 180 so so this right over here has to be a let's see so it's going to be 180 minus 45 minus 30 that's 180 minus 75 so this is going to equal 105 degree angle right over here 105 degree angle and so applying the law of sines here and actually let me label the different sides let's call this side right over here let's call this side right over here side a or has length a and let's call this side right over here let's say this side right over here has length length B so the law of sines tells us that the ratio between the sine of an angle and it the opposite side is going to be constant through this triangle so it tells us that sine of this angle sine of 30 degrees over the length of the side opposite is going to be equal to is going to be equal to sine of 105 degrees sine of 105 degrees over the length of the side opposite to it which is going to be equal to which is going to be equal to sine of 45 degrees sine of 45 degrees equal to the length of the side opposite so sine of 45 degrees over B and so if we wanted to figure out a we could solve this equation right over here and if we wanted to solve for B we could just set this equal to that right over there so let's solve each of these so what is the sine of 30 degrees well you just might remember it from your unit circles or from even 30-60-90 triangles so that's one-half and if you don't remember it you can use a calculator to verify that I already verify that this is in degree mode so it's point five so this is going to be equal to one-half over two so the another way of thinking about it that's going to be equal to one fourth this piece is equal to one fourth is equal to a sine of a hot or is equal to sine of 105 degrees over a so let me write this this is equal to sine of 105 degrees over a and actually we could also say since we could actually kind of do both at the same time that this is equal to that that 1/4 is equal to is equal to sine of 45 degrees over be actually sine of 45 degrees is another one of those that it's easy to jump out of kind of your unit circle you might remember it's square root of 2 over 2 so let's just write that that's square root of 2 square root of 2 over 2 and you could use a calculator but you'll get some decimal value right over there but in either case in either of these equations let's solve for a and then let's solve for B so one thing we could do is we could take the reciprocal of both sides of this equation the reciprocal of 1/4 is 4 and the reciprocal of this right-hand side is a over the sine of 105 degrees and now to solve for a we could just multiply both sides times the sine of 105 degrees so we get 4 times the sine 4 times the sine of 105 degrees is equal to a so let's get our calculator out so 4 times the sine of 105 gives us let's see it's approximately equal to let's say let's just round to the nearest hundreds 3.86 so a is approximately equal to three point eight six so approximately three point eight six which looks about right if this is two and I made my angles appropriately that looks like about 3.86 let's figure out what B is we could once again take the reciprocal of both sides of this and we get four is equal to B over square root of two over two we can multiply both sides times square root of two over two and we would get B is equal to four four times the square root of two over two square root of two over two or we can think of it B is four times the sine of 45 degrees but let's think let's figure out what that is so if we want an actual numerical value we could just write this as two square roots of two but let's actually figure out what that is two square roots of two is equal to two point most is eight three so B B is approximately equal to two point eight three so I'm going to be clear this 4 divided by two is two square roots of two which is two point eight which is approximately equal to approximately equal to two point eight three if we round to the nearest hundreds two point eight three which also seems pretty reasonable here so the key the key of the law of cosines is if you have let's say you have two angles and a side you are able to figure out everything else about it or if you actually had two if you had two sides and an angle you also would be able to figure out everything else about the triangle