Question

在△ABC中,a=4,b+c=5,tanA+tanB=-√3(1-tanA+tanB),求SinA.

Answer 1

tan(A+B)=(tanA+tanB)/(1-tanAtanB)=-√3

tanC=√3 C=60

cosC=(a^2+b^2-c^2)/2ab=0.5

(16+b^2-c^2)/8b=0.5

c=7/2 b=3/2

a/sinA=c/sinC 4/sinA=(7/2)/(√3/2)=4√3/7