# I need the angles of a triangle...forgot how to do it.

I have a triangle the base is 46 inches...the sides are 60 inches. What are the angles....or how do I figure the angles?
Rewrite the law of cosines ( a²=b²+c²-(2bc)(cosA) ) or ( b²=a²+c²-(2ac)(cosB) ) or ( c²=b²+a²-(2ba)(cosC) ) so that you're solving for the angle as opposed to the side. I'll just do it for angle A, you can do it for angles B and C yourself. First we subtract b^2 and c^2 in an attempt to isolate cosA, so we have ( a²-b²-c²=-(2bc)(cosA) ) now we multiply by -1 (we're actually dividing by it, but it gives the same result in this case). ( b²+c²-a²=(2bc)(cosA) ) and then we divide by 2bc giving us ( (b²+c²-a²)/(2bc)=(cosA) ) and finally, we simply do inverse cosine in order to find out which angle equals that value for cosine ( cos^-1((b²+c²-a²)/(2bc))=A ) So that gives you angle A. From here you can use either the other converted cosine rules, or the sine rule to solve for the other angles. And remember, you only need two angles, just add them up and subtract from 180 to get the third.