Question

高一数学问题!!!紧急

在三角形ABC中,ABC成等差数列,sinC=5/13,求cosA

Answer 1

因为ABC成等差数列,所以B=60度,cosB=1/2,sinB=√3/2

sinC=5/13,所以cosC=12/13(这儿不能等于-12/13,否则超过135度,B等于60度,与内角和180度矛盾)
cosA=-cos(B+C)=-(cosBcosC-sinBsinC)=-(12-5√3)/26

Answer 2

∴2B=A+C,A+B+C=2B+B=3B=180°.
B=60°,A+C=120°.
又∵sinC=5/13<1/2,所以C<30°或C>150°(舍去，否则A+B>180°).
∴cosC>0,cosC=12/13
cosA=cos(120°-C)=cos120°*cosC+sin120°*sinC
=(-12+5√3)/26.

Answer 3

A+B+C=180
2B=A+C
B=60

cosA=cos(180-(B+C))=-(cos(B+C))=-(cosBcosC-sinBsinC)=sinBsinC-cosBcosC