1. **State the problem:** Given $\sin A = \frac{8}{17}$$ with $A$ obtuse, and $\cos B = \frac{5}{13}$ with $B$ acute, find $\sin(2A + B)$ without using tables or direct calculation. 2. **Recall formulas and rules:** - Double angle formula for sine: $\sin(2A) = 2 \sin A \cos A$ - Angle addition formula: $\sin(2A + B) = \sin 2A \cos B + \cos 2A \sin B$ - Since $A$ is obtuse ($90^\circ < A < 180^\circ$), $\cos A$ is negative. - Since $B$ is acute ($0^\circ < B < 90^\circ$), $\sin B$ and $\cos B$ are positive. 3. **Find $\cos A$:** Using $\sin^2 A + \cos^2 A = 1$: $$\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \left(\frac{8}{17}\right)^2} = -\sqrt{1 - \frac{64}{289}} = -\sqrt{\frac{225}{289}} = -\frac{15}{17}$$ 4. **Find $\sin B$:** Using $\sin^2 B + \cos^2 B = 1$: $$\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$ 5. **Calculate $\sin 2A$ and $\cos 2A$:** $$\sin 2A = 2 \sin A \cos A = 2 \times \frac{8}{17} \times \left(-\frac{15}{17}\right) = -\frac{240}{289}$$ $$\cos 2A = \cos^2 A - \sin^2 A = \left(-\frac{15}{17}\right)^2 - \left(\frac{8}{17}\right)^2 = \frac{225}{289} - \frac{64}{289} = \frac{161}{289}$$ 6. **Use angle addition formula:** $$\sin(2A + B) = \sin 2A \cos B + \cos 2A \sin B = \left(-\frac{240}{289}\right) \times \frac{5}{13} + \frac{161}{289} \times \frac{12}{13}$$ 7. **Simplify each term:** $$-\frac{240}{289} \times \frac{5}{13} = -\frac{1200}{3757}$$ $$\frac{161}{289} \times \frac{12}{13} = \frac{1932}{3757}$$ 8. **Add the terms:** $$\sin(2A + B) = -\frac{1200}{3757} + \frac{1932}{3757} = \frac{732}{3757}$$ **Final answer:** $$\boxed{\sin(2A + B) = \frac{732}{3757}}$$