In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10, C = √ In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 C = √ 5 for ABC area

In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10, C = √ In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 C = √ 5 for ABC area

cos2A=1-2(sinA)^2=3/5
sinA=√5/5
Because a and B are acute angles
So cosa = 2 √ 5 / 5, CoSb = 3 √ 10 / 10
So sinc = sin (a + b) = sinacosb + cosasinb = √ 2 / 2
Sine theorem a / Sina = C / sinc = √ 10
a=√2
So ABC area s = 1 / 2, acsinb = 1 / 2