0\\\cos x=\sqrt{1-\sin^2x}=\sqrt{1-\frac{9}{13}}=\sqrt{\frac{4}{\sqrt{13}}} =\frac{2}{\sqrt{13}} \\\tg \alpha = \sin \alpha / \cos \alpha = \frac{3}{\sqrt{13}} : \frac{2}{\sqrt{13}}=1.5" alt="\alpha \in (0; \pi/2)\Rightarrow \cos x>0\\\cos x=\sqrt{1-\sin^2x}=\sqrt{1-\frac{9}{13}}=\sqrt{\frac{4}{\sqrt{13}}} =\frac{2}{\sqrt{13}} \\\tg \alpha = \sin \alpha / \cos \alpha = \frac{3}{\sqrt{13}} : \frac{2}{\sqrt{13}}=1.5" align="absmiddle" class="latex-formula">