Hi, I'm new to this and for some reason, the website isn't letting me response properly. Anyways, here's the link to the topic I posted earlier. I still am struggling because I'm not sure how the Law of Cosines apply. If c^2=a^2 +b^2 -2ab(cosC), then how can I apply that? I tried setting the different terms in the inequality to sides of a triangle and then using the law of cosines, but it just gets messy. Any help would be greatly appreciated. https://web2.0calc.com/questions/precalc_121

Saphia1123 Feb 9, 2023

#1**0 **

Since I'm in 9th grade math, all I know is that c^2=a^2+b^2 is the pythagoreon theorom.

Hope that helps in some way!

PausedDuck Feb 9, 2023

#2**0 **

Hey Thank you for the response! Yea I was thinking along the lines of that to help cancel the square root, but I just am not sure how to go from there. I labeled a triangle with c= sqrt(a^2 - ab +b^2) b= sqrt(a^2 - ac +c^2) a= sqrt(b^2 + bc +c^2) Then, I tried putting values into the law of cosines and got: c^2= (b^2 + bc +c^2) + (a^2 - ac +c^2) -2(sqrt(b^2 + bc +c^2)*sqrt(a^2 - ac +c^2)) I don't know how this proves anything. I used the triangle inequality theorem to substitute the terms as a, b, and c in triangle ABC with the lower case letters representing the opposite angles.

Saphia1123
Feb 9, 2023

#3**0 **

Any way to apply the law of cosines? I noticed that the terms of the inequality kinda looks like the law of cosines if you write it like this: a^2-2ab+b^2=cos C. That's difficult because there is a 2ab, and I don't know how to get rid of it. I think maybe involving trying to find the area of a triangle with 1/2 * ab sin angle C. I think I figured out the equality part because it would be a degenerate triangle. But how do I prove the inequality? I know that part is related to the triangle inequality theorem.

Saphia1123
Feb 9, 2023