precalculus
posted by Kimmie .
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.

make two separate diagrams,
one will be the 3,4,5 rightangled triangle,
the other will be the 5,12,13 rightangled 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!