Respuesta :

[tex]\begin{gathered} \sin \theta=\frac{-5}{13} \\ \tan (\pi+\theta)=\tan \theta \\ \sin \theta=\frac{p}{h} \\ \text{base}=\sqrt[]{(hypotonous)^2-(perpendicular)^2} \\ =\sqrt[]{(13)^2-(-5)}^2 \\ =\sqrt[]{169-25} \\ =\sqrt[]{144} \\ =12 \\ \tan \theta=\frac{p}{b} \\ \tan \theta=\frac{-5}{12} \end{gathered}[/tex]

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