Question

Solve triangle ABC given a= 34.5, b= 57.2, and c= 28.8.

Answer 1

what do you need? the angle? the area?

Answer 2

I need help finding the angle's of ABC.

Answer 3

have you covered the Law of Cosines yet?

Answer 4

Yes I have, I just really need help with this question. I understand how to do the problem, but I just want to make sure I have the right answers. Can you please help?

Answer 5

\(\bf \textit{Law of Cosines}\\ \quad \\
c^2 = {\color{blue}{ a}}^2+{\color{red}{ b}}^2-(2{\color{blue}{ a}}{\color{red}{ b}})cos(C)\implies cos(C)=\cfrac{{\color{blue}{ a}}^2+{\color{red}{ b}}^2-c^2}{2{\color{blue}{ a}}{\color{red}{ b}}}\\ \quad \\
\measuredangle C=cos^{-1}\left(\cfrac{{\color{blue}{ a}}^2+{\color{red}{ b}}^2-c^2}{2{\color{blue}{ a}}{\color{red}{ b}}}\right)\\ \quad \\ \quad \\


a^2 = {\color{blue}{ c}}^2+{\color{red}{ b}}^2-(2{\color{blue}{ c}}{\color{red}{ b}})cos(A)\implies
cos(A)=\cfrac{{\color{blue}{ c}}^2+{\color{red}{ b}}^2-a^2}{2{\color{blue}{ c}}{\color{red}{ b}}}\\ \quad \\
\measuredangle A=cos^{-1}\left(\cfrac{{\color{blue}{ c}}^2+{\color{red}{ b}}^2-a^2}{2{\color{blue}{ c}}{\color{red}{ b}}}\right)\)

Answer 6

and angle \(\bf \measuredangle B\) will be 180 -(A+C)

Answer 7

I got 120 degrees for angle B. Is that what you got?

Answer 8

well, I got \(\bf 128.058^o\)

Answer 9

erk rather \(\bf 129.058^o\)

Answer 10

Great, I see how you got that. What did you get for angle A and C? I know the ones I got are wrong. lol