T TrinityHarvin Mar 2019 14 0 United States Mar 20, 2019 #1 Not sure if this will prove as a challenge for any, but these is for me (unfortunately):

T TrinityHarvin Mar 2019 14 0 United States Mar 20, 2019 #3 Short and frank answer, thank you! I could use a little more than just google right now, but thank you!

Short and frank answer, thank you! I could use a little more than just google right now, but thank you!

J joshuaa Mar 2012 564 29 Mar 20, 2019 #4 Let us call the three sides of the triangle A, B, and C. And Θ, the angle we want to find. If A is the opposite side to the angle we want to find then, the cosine law is A^2 = B^2 + C^2 - 2BC cos Θ

Let us call the three sides of the triangle A, B, and C. And Θ, the angle we want to find. If A is the opposite side to the angle we want to find then, the cosine law is A^2 = B^2 + C^2 - 2BC cos Θ

P Plato MHF Helper Aug 2006 22,469 8,640 Mar 20, 2019 #5 TrinityHarvin said: Short and frank answer, thank you! I could use a little more than just google right now, but thank you! Click to expand... O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines. Now the cosine of an obtuse angle is negative. How do you know that? Reactions: 1 person

TrinityHarvin said: Short and frank answer, thank you! I could use a little more than just google right now, but thank you! Click to expand... O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines. Now the cosine of an obtuse angle is negative. How do you know that?

T TrinityHarvin Mar 2019 14 0 United States Mar 20, 2019 #6 So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ?

So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ?

P Plato MHF Helper Aug 2006 22,469 8,640 Mar 20, 2019 #7 TrinityHarvin said: So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ? Click to expand... Because $\cos(\alpha)<0~\&~a^2+2cb\cos(\alpha)=b^2+c^2$ then the answer is $a^2>b^2+c^2$.

TrinityHarvin said: So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ? Click to expand... Because $\cos(\alpha)<0~\&~a^2+2cb\cos(\alpha)=b^2+c^2$ then the answer is $a^2>b^2+c^2$.

J joshuaa Mar 2012 564 29 Mar 20, 2019 #8 TrinityHarvin said: So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ? Click to expand... yes correct and for your triangle above Θ = 36.87°

TrinityHarvin said: So if: A=3 B=4 C=5 And a is the side opposite of Θ, we would substitute the side lengths, leaving it to look something like this: 3^2=4^2+5^2-2(4)(5)cosΘ ? Then we would just solve from there for Θ? Click to expand... yes correct and for your triangle above Θ = 36.87°

J joshuaa Mar 2012 564 29 Mar 20, 2019 #9 Plato said: O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines. Now the cosine of an obtuse angle is negative. How do you know that? Click to expand... I understand what Plato is saying. And I got that situation once, but because I always draw and estimate my solution, I knew the answer was in the wrong Quadrant. So, it was easy to fix the calculator answer and get the required angle!

Plato said: O.K. $a^2+2bc\cos(\alpha)=b^2+c^2$ from the law of cosines. Now the cosine of an obtuse angle is negative. How do you know that? Click to expand... I understand what Plato is saying. And I got that situation once, but because I always draw and estimate my solution, I knew the answer was in the wrong Quadrant. So, it was easy to fix the calculator answer and get the required angle!