Actually, I think the theorem goes....." in any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides"

Originally posted by missal Actually, I think the theorem goes....." in any [b]right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides"[/b]

I was giving you guys credit to know that already. I'll remember not to do that next time.

The full formula from which the Pythagorus Theorem is based is as follows (as I recall)

C^2 = A^2 + B^2 - 2*A*B*cos(o) (where o = the angle between sides A and B)

At 90 degrees, cos(o) = 0, thus eliminating the third term.

At 180 degrees, cos(o) = -1, making the formula C^2 = A^2 + B^2 + 2*A*B, which simplifies into C^2 = (A+B)^2. Given lengths are generally considered positive, C is thus A+B.

Originally posted by geepamoogle The full formula from which the Pythagorus Theorem is based is as follows (as I recall)

C^2 = A^2 + B^2 - 2*A*B*cos(o) (where o = the angle between sides A and B)

At 90 degrees, cos(o) = 0, thus eliminating the third term.

At 180 degrees, cos(o) = -1, making the formula C^2 = A^2 + B^2 + 2*A*B, which simplifies into C^2 = (A+B)^2. Given lengths are generally considered positive, C is thus A+B.