ΔABC的三个内角A,B,C所对的边分别为a,b,c,asinAsinB+bcos^2A=√2a 1.求b⼀a 2.若c2=b2+根号3*a2,求B

2025-06-26 06:48:07
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回答1:

任意三角形,sinA/a=sinB/b,所以asinAsinB+bcos2A(2是次数)=根2a,asinA*(sinA*b)/a=sin2A(2是次数),即bsin2A+bcos2A=根2a,b=根2a,b/a=根2