正弦定理和余弦定理1、在△ABC中,若a:b:c=1:3:5,求2sinA-sinB/sinC的值.2、在△ABC中,C=根号2+根号6,角C=30°,求边a+b的取值范围.
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1、在△ABC中,a:b:c=1:3:5
则由正弦定理a/sinA = b/sinB = c/sinC得：sinA:sinB:sinC = a:b:c = 1:3:5
所以(2sinA-sinB)/sinC = (2*1-3)/5 = -1/5
2、a/sinA=b/sinB=c/sinC=(a+b)/(sinA+sinB);
所以 a+b=c(sinA+sinB)/sinC=2(√6+√2)(sinA+sinB)=4(√6+√2)*sin[(A+B)/2]*cos[(A-B)/2]=4(√6+√2)sin75°*cos[(A-B)/2]=[(√6+√2)^2]*cos[(A-B)/2](其中用到了和差化积的知识)因为 -150°