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
