To evaluate the expression sin(a + B) when sin(a) = -120/169 in Quadrant I and cos(B) = -13 in Quadrant II, we can follow these steps:
1. Since sin(a) = -120/169 in Quadrant I, we can use the Pythagorean identity to find cos(a). The Pythagorean identity states: sin^2(a) + cos^2(a) = 1. Therefore, cos(a) = sqrt(1 - sin^2(a)) = sqrt(1 - (-120/169)^2) = sqrt(1 - 14400/28561) = sqrt(14161/28561) = 119/169.
2. Now that we have sin(a) and cos(a), we can find sin(B) using the fact that cos^2(B) = 1 - sin^2(B). Given that cos(B) = -13 in Quadrant II, sin(B) must be positive. Therefore, sin(B) = sqrt(1 - (-13)^2) = sqrt(1 - 169) = sqrt(-168). Since sin(B) is positive in Quadrant II, sin(B) = sqrt(168).
3. Now, to evaluate sin(a + B), we can use the sum formula: sin(a + B) = sin(a)cos(B) + cos(a)sin(B). Plugging in the values we found earlier, sin(a + B) = (-120/169)(-13) + (119/169)(sqrt(168)) = 1560/169 + 119sqrt(168)/169.
Therefore, the expression sin(a + B) = (1560 + 119sqrt(168))/169.
