a a A B A B SolutionsinB 1213 in QII sinA 35 in QIV cosB

a.) a.) A B) (A B)

sinB = 12/13 in QII ; sinA = -3/5 in QIV

cosB = -sqrt(1 -sin^2B) = -5/13

cosA = sqrt(1 -sin^2A) = 4/5

So, sin(A+B) = sinAcosB +cosAsinB

=(-3/5)(-5/13) + (4/5)(-3/5)

=(15 + 12)/65

=27/65

b) cosA = 1- 2sin^2(A/2)

sin(A/2) = sqrt[(1 -cosA)/2]

= sqrt[ (1 -4/5)/2]

= 1/sqrt10