Trigonometry Obscurity!

[xenogizmo]

Hey everyone,
I was looking over some old pre-calculus exams and I found this rather obscure looking question.. It's about trigonometry.

You're given a triangle ABC, and the legs are a (BC),b (AC), c (AB).
You're given the lenghts of a=5, b-8, and the angle C between them is 140.

The question is, what's the length of the leg "c" and what are the other 2 degrees? :yuck:

Is that even possible?? :bugeye:
Thx,
--Xeno

[devious_]

Are you familiar with the sine and cosine rules?

[xenogizmo]

Of course I am!! :grumpy:
But they can only be applied to a triangle with a right angle, and I tried to divide this triangle to 2 with right angles but it just didnt seem to work out.. Any ideas?

[Hurkyl]

devious is referring to the law of sines and the law of cosines...

Given any triangle with sides A, B, and C with angles \alpha, \beta, \gamma (with A opposite \alpha, etc):


The law of sines:

\frac{\sin \alpha}{A} = \frac{\sin \beta}{B} = \frac{\sin \gamma}{C}


The law of cosines:

C^2 = A^2 + B^2 - 2AB\cos \gamma


(and, of course, similar formulae for the other choices of angle)

[Alkatran]

Notice that since cos(90) = 0, the law of cosine turns into pythagoras' (sp?) theorem with right triangles. That's the rule I assume you were talking about.