Trigonometry Obscurity: Solving Triangle ABC

[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?

Is that even possible??
Thx,
--Xeno

 

[devious_]

Are you familiar with the sine and cosine rules?

 

[xenogizmo]

Of course I am!
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 [itex]\alpha, \beta, \gamma[/itex] (with A opposite [itex]\alpha[/itex], etc):


The law of sines:
[tex]
\frac{\sin \alpha}{A} = \frac{\sin \beta}{B} = \frac{\sin \gamma}{C}
[/tex]

The law of cosines:
[tex]
C^2 = A^2 + B^2 - 2AB\cos \gamma
[/tex]

(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.