在三角形abc中，cosA=5分之2倍根5，tanB=3分之1。求角c的大小。若三角形abc的外接圆半径为1，求三角形a

(1) cosA = 2√5/5 > 0, A为锐角; sinA = √(1 - cos²A) = √(1 - 1/5) = √5/5 tanB = 1/3 > 0, B为锐角; sinB/√(1 - sin²B) =1/ 3, sinB = √10/10, cosB = sinB/tanB = 3√10/10 cosC = cos[π-(A+B)] = -cos(A+B) = -cosA*cosB + sinA*sinB = -(√5/5)(√10/10) + (2√5/5)(3√10/10) = -√2/10 + 3√2/5 = √2/2 C = π/4 (2) 正玄定理 a/sinA=b/sinB=2 a=2sinA=2√5/5 b=2sinB=6√10/10 S=1/2absinC=3/5