2. Дано: $$\cos \alpha = -\frac{8}{13}$$, $$\frac{\pi}{2} < \alpha < \pi$$. Найти: $$\sin \alpha$$, $$\operatorname{ctg} \alpha$$.

$$\sin^2 \alpha + \cos^2 \alpha = 1$$

$$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(-\frac{8}{13}\right)^2 = 1 - \frac{64}{169} = \frac{169 - 64}{169} = \frac{105}{169}$$

Так как $$\frac{\pi}{2} < \alpha < \pi$$, то $$\sin \alpha > 0$$.

$$\sin \alpha = \sqrt{\frac{105}{169}} = \frac{\sqrt{105}}{13}$$

$$\operatorname{ctg} \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{-\frac{8}{13}}{\frac{\sqrt{105}}{13}} = -\frac{8}{\sqrt{105}} = -\frac{8 \sqrt{105}}{105}$$

Ответ: $$\sin \alpha = \frac{\sqrt{105}}{13}$$, $$\operatorname{ctg} \alpha = -\frac{8 \sqrt{105}}{105}$$