This is a discussion on Please help me with a math (trig) assignment: finding angle within the A Brief History of Cprogramming.com forums, part of the Community Boards category; I have to find the measure of the largest angle in this triagnle. The triangle has sides with a measure ...

I have to find the measure of the largest angle in this triagnle. The triangle has sides with a measure of 121m, 173m, and 194m. It is not a right triangle because 121^2 + 173^2 != 194^2. I am not given any angles. I know the sine law, but I am not sure what steps I should take from here on out. If anyone could 'ask me the right questions' to show me how to get back on track I'd highly appreciate it.

Here's a picture of what I have so far:

EDIT:
sinC = h / 121
sinA = h / 121
194sinA = 121sinC (these both solve for h)
x^2 + h ^2 = 121^2
(173 - x) ^2 + h ^2 = 194 ^2
I really can't think of anything else. Is this even possible?

2. if you use cosine instead of sine, and the cosine law, it might work out.

EDIT: i got
A .= 38
B .= 93 <==
C .= 49

3. use the law of cosines

cos C = (a^2 + b^2 - c^2)/2ab

that will give u the measure of angle C, then u can use the law of sine to find the other angles:

sin A/a = sin B/b = sin C/c

4. yea it should work out for you...cosine law should word fine...

remember that c^2 = h^2 +(173-x)^2

and that c^2 = a^2 + b^2 -28(a)(173)cosC...I think...

...let us know if you have any problems...I might not be able to code much but math I can do...

...but I think the guys have you covered here...

5. In the future, you can try here: Clicky

There's quite a few math people there and they're willing to help.

6. Ok thanks you guys. I haven't been taught the law of cosines yet so I dont' know why they've given us this problem, but i'll look into solving it

7. Hi commander, I actually got a different answer, and I'm not sure who is right. I used both of these following equations ( the first is how I would've done it on my own and the second is what was shown to me in this thread, but they're the same)

cosC = (c^2 - (a^2 + b^2)) / -2(ab)
cosC = (c^2 + b^2 - c^2) / 2ab

here is what I got for angles:
angle C = 80.46
angle B = 61.58
angle A = 37.96

EDIT: I'm right I think, look at this site, move every one of my angles in the I posted picture above clockwise to get the same placement they did

type in 194 for a, 121 for b, and 173 for c, then click angle C and you should get 61.57

http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html