A = 2, CoSb = 3 / 5, C = 45 ° to find the area of triangle ABC Solving problems with cosine theorem

A = 2, CoSb = 3 / 5, C = 45 ° to find the area of triangle ABC Solving problems with cosine theorem

From the title:
cosC=√2/2
cosB=(a*a+c*c-b*b)/2ac=3/5
cosC=(a*a+b*b-c*c)/2ab=√2/2
The results are as follows
20+5*c*c-5*b*b=12c
8+2*b*b-2*c*c=4√2*b
So:
b=8√2/7,c=10/7
Area s = 1 / 2 * a * b * sinc = 8 / 7