Question

在三角形ABC中.cosB等于负十三分之五.cosC等于五分之四.求cosA的值

Answer 1

cosB等于负十三分之五,那么sinB=12/13
cosC等于五分之四,那么sinC=3/5
cosA
=cos[180°-(B+C)]
=-cos(B+C)
=-cosBcosC+sinBsinC
=-(-5/13)×4/5+12/13×3/5
=20/65+36/65
=56/65