jmc

We have that $\sin 70^{\circ }=\cos 20^{\circ },$ $\sin 260^{\circ }=-\sin 80^{\circ }=-\cos 10^{\circ },$ and $\cos 280^{\circ }=\cos 80^{\circ }=\sin 10^{\circ },$ so $$\sin 70^{\circ }\cos 50^{\circ }+\sin 260^{\circ }\cos 280^{\circ }=\cos 20^{\circ }\cos 50^{\circ }-\sin 10^{\circ }\cos 10^{\circ }.$$Then by product-to-sum, $$\begin{array}{rl}\cos 20^{\circ }\cos 50^{\circ }-\sin 10^{\circ }\cos 10^{\circ }& =\frac{1}{2}(\cos 70^{\circ }+\cos 30^{\circ })-\frac{1}{2}\cdot 2\sin 10^{\circ }\cos 10^{\circ }\\ & =\frac{1}{2}\cos 70^{\circ }+\frac{1}{2}\cos 30^{\circ }-\frac{1}{2}\sin 20^{\circ }\\ & =\frac{1}{2}\cos 30^{\circ }=\boxed{\frac{\sqrt{3}}{4}}.\end{array}$$