The problem can be solved using the cosine rule, where cos(A)=(b^2+c^2-a^2)/(2bc) =(12.5^2+12.5^2-8^2)/(2*12.5*12.5) =.7952 => A=2x=37.33 deg. => x=18.66 deg.
HOWEVER, since the triangle is isosceles, we can divide it into two right triangles, with short leg=4", and hypotenuse 12.5", and solve easily by x=asin(4/12.5)=18.66 degrees
