Let sin a = 12 13 with a in qii and sin b = − 15 17 with b in qiii. find sin(a + b), cos(a + b), and tan(a + b)

Sin(A) = 15/17 15^2 + x^2 = 17^2 17^2 - 15^2 = x^2 289 - 225 = x^2 64 = x^2 8 = x 
cos(A) = -8/17 
 
QIII sin(B) = -4/5 (-4)^2 + x^2 = 5^2 5^2 - (-4)^2 = x^2 25 - 16 = x^2 9 = x^2 3 = x 
cos(B) = -3/5  knowing the identity ; sin(A+B) = sin(A) cos(B) + cos(A) sin(B) sin(A+B) = (15/17) (-3/5) + (-8/17) (-4/5) sin(A+B) = (-9/17) + (32/85) sin(A+B) = (-13/85) 

knowing the identity ; cos(A+B)=cos(A) cos(B) - sin(A) sin(B) cos(A+B) = (-8/17) (-3/5) - (15/17) (-4/5) cos(A+B) = (24/85) + (12/17) cos(A+B) = (84/85) 

knowing the identity ; tan(A+B) = sin(A + B) / cos(A + B) tan(A+B) = (-13/85) / (84/85) tan(A+B) = (-13/84)

It is 2 out of 8 she will pick green and 3 out of 7 she will pick green but red and green would be 5 out of 5 so 50% chance