Question

在△ABC和△A1B1C1中,满足sinA=cosA1,sinB=cosB1,sinC=cosC1

1.求证△ABC是钝角三角，并求最大角的度数。
2.求sin^2A+sin^2B+sin^2C的最小值。

Answer 1

(1)假如A,B,C都为锐角,A1=pi/2-A,B1=pi/2-B,C1=pi/2-C
三式相加A1+B1+C1=3pi/2与三角形内角和为180矛盾故有个角为钝角，最大角不能判定，也不能判定度数
(2)(sinA)^2+(sinB)^2+(sinC)^2=(1-cos2A)/2+(1-cos2B)/2+(1-cos2C)/2=3/2-1/2(cos2A+cos2B+cos2C)-(cos2A+cos2B+cos2C)<=-3cos((2A+2b+2c)/3)=3/2所以(sinA)^2+(sinB)^2+(sinC)^2<=3/2+3/4=9/4