DESPERATELY NEED HELP. GEOMETRY. COSINES/SINES. FIND C AND ROUND TO THE NEAREST TENTH

Don't get desperate Erica. These aren't so bad. I went over the Trig menu last time; let's talk about when we might use Law of Sines, Law of Cosines, or Triangle Angles.
If we have two angles we could use Triangle Angles (they add to 180) to get the third.
The Law of Sines takes two angles and their opposite sides; we use it when we have three of four of those.
The Law of Cosines takes three sides and one angle; we use it when we have three of four of those.
Pro Tip: If you ever get a choice between the Law of Cosines and the Law of Sines, use the Law of Cosines.
This one we only have sides, so it's Law of Cosines time to get the angle.
It's important to label the sides correctly. c=16 feet opposite vertex C. a=17 and b=18 (but it wouldn't be a problem to swap those.)
[tex]c^2 = a^2 + b^2 - 2 a b \cos C[/tex]
[tex]2 ab \cos C = a^2 + b^2 - c^2[/tex]
[tex]\cos C = \dfrac{a^2 + b^2 - c^2}{2 ab}[/tex]
[tex]C = \arccos \dfrac{a^2 + b^2 - c^2}{2 ab}[/tex]
The cosine is great for triangle angles because a given cosine uniquely determines a triangle angle between 0 and 180 degrees.
[tex]C = \arccos \dfrac{17^2 + 18^2 - 16^2}{2 (17)(18)} = \arccos \dfrac{7}{12}\approx 54.31^\circ [/tex]
Answer: 51.3°
I think its 45. 8 2 + 16 2 = about 17 2
That means its a right triangle. So my guess is 45.