已知三角ABC,角a是15度,角b是90,角c是75度,bc为7厘米,求ac和ab的长度...答:c^2=7/cos15度, c=2.692 ,b=2.600 ,a=c*cos15度=2.821 ab=7.334 ,ac=7.594,1,正弦定理啊~~a/sinA=b/sinB=c/sinC,1,a/sinA=b/sinB=c/sinC sin15=sin(45-30)=sin45cos30-sin30cos45=(根号6-根号2)/4 得b=7(根号6+根号2) 由勾股定理求c.,1,AB=14+7√3,AC=7...