$\sin ^{2} 10^{\circ}+\cos ^{2} 40^{\circ}+\sin 10^{\circ} \cos 40^{\circ}=$ ________.
答案 $\dfrac 34$．
解析 在 $\triangle ABC$ 中, 设 $A,B,C$ 分别为 $10^\circ,50^\circ,120^\circ$, 则根据正弦定理, 有\[\dfrac{a}{\sin10^\circ}=\dfrac{b}{\sin 50^\circ}=\dfrac{c}{\sin120^\circ}=d,\]于是\[\sin ^{2} 10^{\circ}+\cos ^{2} 40^{\circ}+\sin 10^{\circ} \cos 40^{\circ}=\dfrac{a^2+b^2-2ab\cos C}{d^2}={c^2}{d^2}=\sin^2120^\circ=\dfrac 34.\]