i) Chamarei os ângulos dos vértices $\widehat A$, $\widehat B$ e $\widehat C$ de $\alpha$, $\omega$ e $\theta$, respectivamente. Como temos um triângulo, note que $\alpha + \omega + \theta = 180 \therefore \\ \alpha = 180 - (\omega + \theta) \\ \omega = 180 - (\alpha + \theta) \\ \theta = 180 - (\alpha + \omega)$ ii) $sen(\omega) = sen(180 - (\alpha + \theta)) = sen(\alpha + \theta)$ $sen(\theta) = sen(180 - (\alpha + \omega)) = \frac{4}{5} \therefore cos(\theta) = \frac{3}{5}$ iii) $sen(\alpha - \theta) = \frac{4}{5} \therefore 3sen(\alpha) - 4cos(\alpha) = 4$ $(3sen(\alpha))^2 = (4 + 4cos(\alpha))^2$ $9sen^2\alpha = 16 + 32cos(\alpha) + 16cos^2\alpha \Rightarrow 9(1-cos^2\alpha) = 16cos^2\alpha + 32cos(\alpha) + 16$ $25cos^2\alpha + 32cos(\alpha) + 7 = 0 \therefore cos(\alpha) = -\frac{7}{25} \therefore sen(\alpha) = \frac{24}{25}$ iv) $sen(\omega) = sen(\alpha + \theta) \Leftrightarrow sen(\omega) = \frac{24}{25}.\frac{3}{5} + \frac{4}{5}.\frac{-7}{25} \therefore sen(\omega) = \frac{44}{125}$