Question 1201490
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sin(B)=5/13 in quadrant II, so cos(B)=-12/13. (5-12-13 is a Pythagorean Triple)<br>
The point (6,-8) is on the terminal side of angle A, so angle A is in quadrant IV, and tan(A)=-4/3.  That makes sin(A)=-4/5 and cos(A)=3/5.<br>
sin(A+B) = sin(A)cos(B)+cos(A)sin(B) = (-4/5)(-12/13)+(3/5)(5/13) = 48/65+15/65 = 63/65.<br>
ANSWER: 63/65<br>