Question

在三角形ABC中,COSA=负十三分之五,COSB=五分之三

Answer 1

sin (C)

=sin (π-A-B)

=sin (A+B)

=sin (arccos (-5/13)+arccos (3/5) )

=sin (arccos (-5/13) )cos (arccos (3/5) )+cos (arccos (-5/13) )sin (arccos (3/5) )

=3/5 √(1-5^2/(13)^2 )-5/13 √(1-3^2/5^2 )

=3/5⋅12/13-5/13⋅4/5

=16/65

While BC=a=5,

b=(a sin B)/sin A

=(a√(1-cos^2 B ))/√(1-cos^2 A )

=5⋅4/5⋅13/12

=13/3

Area=1/2 ab sin C

=1/2⋅5⋅13/3⋅16/65

=8/3