$\left(2 \sqrt{3} \sin 70^{\circ}-\tan 70^{\circ}\right) \sin 80^{\circ}=$
解三角形; 2.三角恒等变换.

14. 答案:$\frac{\sqrt{3}}{2}$

【解析】

$\because \alpha 、 \beta$ 为锐角, $\cos \alpha=\frac{1}{7}, \therefore \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{4 \sqrt{3}}{7}$,

$\sin (\alpha+\beta)=\frac{5 \sqrt{3}}{14}<\sin \alpha, \therefore \alpha+\beta$ 为针角,

$\therefore \cos (\alpha+\beta)=-\sqrt{1-\sin ^{2}(\alpha+\beta)}=-\frac{11}{14}$,

$\therefore \sin \beta=\sin [(\alpha+\beta)-\alpha]=\sin (\alpha+\beta) \cos \alpha-\cos (\alpha+\beta) \sin \alpha$

$=\frac{5 \sqrt{3}}{14} \cdot \frac{1}{7}-\left(-\frac{11}{14}\right) \cdot \frac{4 \sqrt{3}}{7}=\frac{\sqrt{3}}{2}$.

故答案为 $\frac{\sqrt{3}}{2}$.