Cos A Plus C F92221

1. **State the problem:** We are given a triangle $\triangle ABC$ with the ratio of sines of its angles as $\sin A : \sin B : \sin C = 4 : 5 : 6$. We need to find $\cos(A + C)$. 2. **Recall the sine rule:** In any triangle, the sides are proportional to the sines of the opposite angles: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$$ where $a,b,c$ are the sides opposite to angles $A,B,C$ respectively. 3. **Use the given ratio:** Since $\sin A : \sin B : \sin C = 4 : 5 : 6$, we can write $$\sin A = 4x, \quad \sin B = 5x, \quad \sin C = 6x$$ for some positive $x$. 4. **Sum of angles in a triangle:** We know $$A + B + C = 180^\circ$$ which implies $$A + C = 180^\circ - B$$ 5. **Express $\cos(A + C)$:** Using the above, $$\cos(A + C) = \cos(180^\circ - B) = -\cos B$$ 6. **Find $\cos B$ using $\sin B$:** Since $\sin B = 5x$, and $\sin^2 B + \cos^2 B = 1$, $$\cos B = \pm \sqrt{1 - \sin^2 B} = \pm \sqrt{1 - (5x)^2} = \pm \sqrt{1 - 25x^2}$$ 7. **Determine $x$:** Since $\sin C = 6x \leq 1$, $$6x \leq 1 \implies x \leq \frac{1}{6}$$ 8. **Check if $x=\frac{1}{6}$ works:** $$\sin A = 4 \times \frac{1}{6} = \frac{2}{3} \approx 0.6667$$ $$\sin B = 5 \times \frac{1}{6} = \frac{5}{6} \approx 0.8333$$ $$\sin C = 6 \times \frac{1}{6} = 1$$ $\sin C = 1$ means $C = 90^\circ$. 9. **Find $B$ and $A$:** Since $C=90^\circ$, $$A + B = 90^\circ$$ 10. **Find $B$ from $\sin B = \frac{5}{6}$:** $$B = \arcsin\left(\frac{5}{6}\right) \approx 56.44^\circ$$ 11. **Find $A$:** $$A = 90^\circ - B \approx 33.56^\circ$$ 12. **Calculate $\cos B$:** $$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{5}{6}\right)^2} = \sqrt{1 - \frac{25}{36}} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6}$$ 13. **Calculate $\cos(A + C)$:** $$\cos(A + C) = -\cos B = -\frac{\sqrt{11}}{6}$$ **Final answer:** $$\boxed{\cos(A + C) = -\frac{\sqrt{11}}{6}}$$