Question

在角ABC中,tanB=1,tanC=2,b=100,求边c

Answer 1

tanA

= tan(180-B-C)

= -tan(B+C)

= -[(tanB+tanC)/(1-tanB*tanC)]

= -3

1/(cosA)^2

=(secA)^2

=(tanA)^2+1

=10

所以(cosA)^2=1/10

(sinA)^2=9/10

由tanB>0 tanC>0,tanA