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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 What's the math problem 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 Find the value of a + B?
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 COSC = - cos (a + b) = - cosacosb + sinasinb = - √ 2 / 2
So C = 3pi / 4
So a + B = pi-3pi / 4 = pi / 4
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