Cosign Formula Field with airSlate SignNow

Cosign formula field

let's say side a is 10 and side B is 20 and angle C is 60 degrees go ahead and solve the triangle so first let's draw it so this is angle a B and C so angle C is 60 degrees side a is 10 side B is 20 so what we have is a side angle side triangle can we use law of sines to solve the triangle in order to use law of sines you need to have two of the same letter notice that we can't use it we have one of each different letter so in this case we need to use the law of cosines if you try to use the law of sines you're gonna miss something for example let's say if we try to use a over sign a which is equal to B over sine B it's not gonna work we have a and B but we don't have angle a nor do we know angle B and if we try to use B over sine B which is equal to C over sine C we're still missing angle B and we're missing side C so whenever you have all different letters you cannot use the law of sines to solve it however there's something else that we can use and that is the law of cosines so here's the formula that you need C squared is equal to a squared plus B squared minus 2a B cosine of angle C now you can change it up and write two of the forms a squared is equal to B squared plus c squared minus 2bc cosine of angle a or b squared is equal to a squared plus c squared - to AC cosine of angle B you can use any one of these three forms but I'm going to use the first one because we have everything to use that formula we have side a and B a is 10 and B is 20 and we have angle C which is 60 degrees so we can use this to find side C 10 squared is 100 20 times 20 is 400 and 2 times 10 is 20 times another 20 that's 400 as well now cosine 60 is one half a hundred plus 400 is 500 and half of 400 is 200 so C squared is equal to 300 so therefore C is the square root of 300 which is 10 root 3 or 17 point 32 so now that we have side C we could use the law of sines to figure out everything else so let's use the law of sines to find angle B C over sine C is equal to B over sine B side C is seventeen point three two and angle C is 30 B is 20 let's go ahead and find angle B so let's cross multiply so 20 times sine 60 that's 17 point 32 and that's equal to 17 point 32 sine B so if we divide both sides by 17.3 2 what that means is that sine B is equal to 1 so therefore B is the arc sine of 1 which is 90 degrees so there's only one answer here because if you do 180 minus 90 you're gonna get 90 again now to find angle a that's gonna be 180 minus 90 minus 60 which is going to be 30 and so that's how you can solve this particular triangle it turns out that it's a right triangle let's say that side a is seven side B is eight side C is nine use a law of cosines to solve the triangle so this time we have all three sides so what we have is a side side side triangle so let's start with this formula C squared is equal to a squared plus B squared minus 2a B cosine of angle C so first let's subtract both sides by a squared and B squared if we move it to the left side we're gonna have C squared minus a squared minus B squared is equal to negative 2a B cosine of C now let's divide both sides by negative two a B so this is the formula we're gonna use cosine of angle C is equal to C squared minus a squared minus B squared divided by negative two a B C is 9 a is 7 and B is 8 9 squared is 81 7 squared is 49 and 8 squared is 64 so 81 minus 49 minus 64 that's equal to negative 32 and then 2 times 7 times 8 that's 112 so negative 32 divided by a negative 1 12 that's 2 over 7 which as a decimal is point two eight five seven that's equal to cosine of angle C so angle C is going to be the arc cosine of that number point two eight five seven so you should get about seventy three point three degrees actually 0.4 for angle seen now let's use the law of cosines to figure out everything else so let's start with this equation C over sine C is equal to a over sine a so C is 9 angle C is 73 point for side a is 7 now let's cross multiply 7 times sine seventy three point four that's about six point seven zero eight and that's equal to nine times sine a so six point seven zero eight divided by nine that's about point seven four five three so angle a is going to be the arc sine of that number so that's 48 point two degrees now let's find the other answer so a could be 180 minus 48 point two which is 131 point eight but this answer is not possible because if we added this to the pre-existent angle and that exceeds 180 so therefore there's only one possible solution one triangle that can be formed so a let's write the answer that's a forty eight point two degrees and now let's calculate angle B so B is going to be 180 minus seventy three point four minus forty eight point two and that's a fifty eight point four degrees so if you have all three sides you need to use the law of cosines to find a first angle and then you can use the law of sines to find everything else so that's it for this lesson