this is a cool website... http://ostermiller.org/calc/triangle.html. but i still wanna know the formula.

If I know it from memory or have to figure it out. Triangle angles = 180* find the ratio of the sides to each other and divide 180* by those ratios. Ratios with 1-1-2 the angles = 45*45*90*. I'm not a teacher, that the best I can explain it.

Sin(x)=opposite/ hypotenuse Cos(x)= adjacent/ hypotenuse Tan(x)= opp/adj Inverse of the above, when the appropriate lengths for the trig function are entered should give you the value of x. sin^-1(opp/hyp)=x and so on. Right? its been a while since trig.

trig doesn't only work on right triangles, but the equations posted do only work on right triangles. the only trig function i know that does work on non-right triangles is: a/sin(A)=b/sin(B)=c/sin(C).

You need to use the law of cosines. a^2 = b^2 + c^2 - ((2ab)cos(C)) Once you find (C), work from there EDIT- found a nice applet: http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html