Law of Sines Part 2 (If there is a Part 1 I Don't Remember but I'm Sure I Made One B4)

Ok so, I'm going over law of sines again. Law of sines can only be used on non-right triangles. The formulas you use for this are: SinA/a = SinB/b = SinC/c

Say you have a Triangle ABC, which A=60degrees, a=4, b=3; you are told to solve the triangle. In a case like this, when you have two legs and at least one angle (of course, the angle has to be relevant to one of the legs), you would use law of sines to solve the triangle.

First thing you would do is set up the formula, so Sin60/4 = SinB/3, cross multiply using your calculator (3Sin60/Sin4); get B = 37.245
Now add B and A and subtract that number from 180 to find angle C and get 82.755.
Use law of sines to find c. Sine82.755/c = Sine60/4; get 4.582, so you end up with:
A= 60, B = 37.245, C = 82.755, a = 4, b = 3; c = 4.582

~ Parrish Masters