i) tomando o triângulo em questão, pela lei dos senos teremos que: $\frac{p}{sin(\hat{P})} = \frac{q}{sin(\hat{Q})} = \frac{r}{sin(\hat{R})} = 2R $ $\therefore \ p =2R.sin(\hat{P})$ $\therefore \ q =2R.sin(\hat{Q})$ $\therefore \ r =2R.sin(\hat{R})$ ii) reescrevendo a matriz e calculando seu deteterminante, teremos: $\begin{vmatrix} 1 & 1 & 1\\ 2R.sin(\hat{P})&2R.sin(\hat{Q}) & 2R.sin(\hat{R})\\ sin(\hat{P}) & sin(\hat{Q}) & sin(\hat{R}) \end{vmatrix}$ $= 2R.sin(\hat{Q}).sin(\hat{R}) + 2R.sin(\hat{R}).sin(\hat{P}) + 2R.sin(\hat{P}).sin(\hat{Q}) - \\ (2R.sin(\hat{P}).sin(\hat{R}) + 2R.sin(\hat{R}).sin(\hat{Q})) + 2R.sin(\hat{Q}).sin(\hat{P}))$ $= 0$