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 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!

To find the value of sin(a + b), we need to know the values of both a and b.

In the given information, we are given the value of sin a and cos B, but not the values of a and b themselves.

To find the value of sin(a + b), we can use the following trigonometric identity: sin(a + b) = sin a * cos b + cos a * sin b.

To find the value of sin a, we are given that sin a = 3/5 when Pi/2 < a < Pi. This means that in the second quadrant, the ratio of the opposite side to the hypotenuse is 3/5.

To find the value of cos B, we are given that cos B = -12/13 when Pi < B < 3Pi/2. This means that in the third quadrant, the ratio of the adjacent side to the hypotenuse is -12/13.

However, without additional information about the specific values of a and b, we cannot calculate the exact value of sin(a + b).

If you have more information about the angles a and b, please provide it so that we can proceed with the calculation.