Question

三角形ABC中.（a+b+c）（a-b+c）=ac,若sinAsinC=（根号3-1）／4 求角C

Answer 1

(a+c+b)(a+c-b)=ac

(a+c)²-b²=ac

a²+c²-b²=-ac

cosB=(a²+c²-b²)/(2ac)=-1/2,B=120°.

sinAsinC= (√3-1)/4

sin(60°-C)sinC=(√3-1)/4.

√3cosCsinC/2-sin²C/2=(√3-1)/4

√3sin(2C)/4-[1-cos(2c)]/4=(√3-1)/4

sin(2C+30°)=√3/2

C=15°或45°.