2. Calculate $\sin ^{2} 20^{\circ}+\cos ^{2} 50^{\circ}+\sin 20^{\circ} \cos 50^{\circ}=$
In $\triangle A B C$, $\sin ^{2} C=\sin ^{2} A+\sin ^{2} B-2 \sin A \sin B \cos C$. Taking $A=20^{\circ}, B=40^{\circ}$, we get $\sin ^{2} 20^{\circ}+\sin ^{2} 40^{\circ}-2 \sin 20^{\circ} \sin 40^{\circ} \cos 120^{\circ}=\sin ^{2} 120^{\circ}=\frac{3}{4}$, i.e., the original expression $=\frac{3}{4}$.