Question

【赏】三角形ABC中,cos(A-C)+cosB=3/2,b^2=ac,求角B.

Answer 1

cos(A-C)+cosB=cos(A-C)-cos(A+C)=cosAcosC+sinAsinC-cosAcosC+sinAsinC

=2sinAsinC=3/2

sinAsinC=3/4

根据正弦定理,a/sinA=b/sinB=c/sinC=2R

b^2=sin^B*4R^2 a=sinA*2R c=sinC*2R

所以,sin^B=sinA*sinC=3/4

因为B