已知 $\alpha \in{ }^{\left(\frac{\pi}{2}, \pi\right)}, \tan \alpha=-\frac{3}{4}$, 则 $\sin \alpha$ 为 $(\square)$
$\because \alpha \in^{\left(\frac{\pi}{2}, \pi\right)}$, $\tan \alpha=-\frac{3}{4}=\frac{\sin \alpha}{\cos \alpha}$, 且 $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \sin \alpha>0, \cos \alpha<0, \therefore \sin \alpha=\frac{3}{5}$.