CAS is good for checking for mistakes/errors.
So, two angles and, knowing the area, perhaps the Hedron formula (which, I never really knew existed)?
To solve for a triangle, you need 3 independent pieces of information -
3 sides, 2 sides + 1 angle, 1 side + 2 angles, but not 3 angles, because the third one becomes redundant.
2 sides + area also works because Heron's formula gives you the third side.
I'm not sure if 2 angles + area works. By this logic it should, but I can't think of how.
Infinitely many triangles with the same angles and area? I don't think so. Give me an example of 2 triangles with the same internal angles and area, but different sides. You can't.
Law of Cos gives you a third side. Heron's(which I just remembered isn't what I was using) depends on the length of the sides. The formula I used has no name, I got it out of my precal book. It works, there no reason debating it until you give an example where it doesn't.
>> I'm not sure if 2 angles + area works
Given 2 angles, you can always derive the 3rd. Just subtract the 2 from 180... From 3 angles and area, just use my formulas.
>> To solve for a triangle, you need 3 independent pieces of information -
3 angles, and area. area being dependent on angles and side length.
Last edited by User Name:; 11-14-2010 at 05:33 PM.
That's why I said it should be possible to solve for the triangle, but I don't know how.
I just showed you how!
Lowercase being sides and uppercase being their opposite angles.Code:a = sqrt((2*sinA*K)/(sinB*sinC)) b = sqrt((2*sinB*K)/(sinA*sinC)) c = sqrt((2*sinC*K)/(sinA*sinB))
Ah! ok. I'll admit I only skimmed the thread .