According to the formula a/sinA = b/sinB = c/sinC = 2r (where r is the radius of the circumscribed circle of the triangle).

r = √2 .sinA= a/2r sinC = c/2r sinB = b/2r

(1) 2 √ 2 (Sina * Sina-sinc * sinc) = 2 √ 2 [(a/2r) * (a/2r)-(c/2r)]

= (a*a-c*c)/2√2

(2)(a-b)sinb =(a-b)* b/2r =(a * b-b * b)/2√2

Because the above two formulas are equal, a*a -c*c =a*b -b*b, so a * a+b * b-c * c = a * b.

According to the cosine theorem, substitute CosC = (a*a+b*b-c*c)/ 2ab into the above formula to get cosC = 1/2.

So c = 60.

(2) Q:

Triangle area s = 0.5a * bsinc = 0.25ab

If and only if a = b, the above formulas are equal, that is, the triangle is an equilateral triangle, and because the radius of the circumscribed circle of the triangle is √2, so

The side length of a triangle is a=b=c=2rsinA = √3*√2 =√6.

The maximum value of s is 0.25 * 0.25 √ 3 (√ 6+√ 6) =1.5 √ 3.

This problem is a comprehensive one. To do such a problem, we must first flexibly judge what formula to use according to the conditions. Since the radius of the circumscribed circle of a triangle is provided, a/sinA = b/sinB = c/sinC = 2r can be used.

There is also a condition of 2 √ 2 (Sina * Sina -SINC * SINC) = (a-b) SINB, which has both angles and sides, so it is unified as angles.

Then we come across the form of a*a +b*b-c*c=a*b, so we use the cosine theorem CosC = (a*a+b*b-c*c)/ 2ab, transform it and skillfully calculate c = 60.

The following topic for calculating triangles can only determine one angle c, so our previous formula for calculating triangles is not easy to use, so we will use the formula S = 0.5absinC to calculate, so we can use the mean inequality (4 ab