Question

在△ABC中,abc分别是角ABC的对边,且cos(B)/cos(c)=-b/(2a+c),求角B？

Answer 1

cos(B)/cos(c)=-b/(2a+c),
(2a+c)*cosB=-b*cosC,
(2a+c)*(a^2+c^2-b^2)/2ac=-b*(a^2+b^2-c^2)/2ab,
化简得,
a^2+c^2-b^2=-ca,
cosB=(a^2+c^2-b^2)/2ac=-1/2=cos120,
B=120度,
2.b=√13,a+c=4.
(a+c)^2=16,
a^2+b^2=16-2ac,
而,cosB=-1/2=(a^2+c^2-b^2)/2ac,
-ac=a^2+c^2-13,
-ac=16-2ac-13,
ac=3/2.
S-三角形面积=1/2*sinB*ac=1/2*√3/2*3/2
=(3√3)/8.,4,在△ABC中,abc分别是角ABC的对边,且cos(B)/cos(c)=-b/(2a+c),求角B
若b=√13，a+c=4。求三角形面积