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If sin a = 3/5 when Pi/2<a<Pi and

cos B = -12/13 when Pi<B<3Pi/2, then what is the value of sin(a+b)?

I don't get the question... Help appreciated.

2 answers

make two separate diagrams,
one will be the 3,4,5 right-angled triangle,
the other will be the 5,12,13 right-angled triangle

given sin a = 3/5 and a is in the II quadrant, so cos a = -4/5

given cos B = -12/13 and B is in III, so
sin B = - 5/13

now you have to know the expansion for
sin(a+b) = sina cosb + cosa sinb
=(3/5)(-12/13) + (-4/5)(-5/13) = -16/65
Thanks!