Problemas de desafío: longitud de arco (radianes) 2

$\begin{array}{rl}\text{Circunferencia}& =2\pi r\\ & =2\pi (30)\\ & =60\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{longitud del arco}}{\text{circunferencia}}}& ={\displaystyle \frac{\text{medida del arco}}{\text{radianes en un círculo}}}\\ \\ \\ {\displaystyle \frac{13\pi }{60\pi }}& ={\displaystyle \frac{\text{medida del arco}}{2\pi }}\\ \\ \\ \text{medida del arco}& =2\pi \cdot {\displaystyle \frac{13\pi }{60\pi }}\\ \\ \\ & ={\displaystyle \frac{13}{30}}\pi \end{array}$‍

$\begin{array}{r}{\displaystyle \frac{13}{30}}\pi +\text{m}\mathrm{\angle }APB=\pi \\ \text{m}\mathrm{\angle }APB={\displaystyle \frac{17}{30}}\pi \end{array}$‍

$\begin{array}{rl}\text{Circunferencia}& =2\pi r\\ & =2\pi (27)\\ & =54\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{longitud del arco}}{\text{circunferencia}}}& ={\displaystyle \frac{\text{medida del arco}}{\text{radianes en un círculo}}}\\ \\ \\ {\displaystyle \frac{\left({\displaystyle \frac{81}{10}}\pi \right)}{54\pi }}& ={\displaystyle \frac{\text{medida del arco}}{2\pi }}\\ \\ \\ \text{medida del arco}& ={\displaystyle \frac{81}{10}}\pi \cdot {\displaystyle \frac{2\pi }{54\pi }}\\ \\ \\ & ={\displaystyle \frac{3}{10}}\pi \end{array}$‍

$\begin{array}{r}{\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{3}{10}}\pi +\text{m}\mathrm{\angle }CPD=\pi \\ \text{m}\mathrm{\angle }CPD={\displaystyle \frac{\pi }{5}}\end{array}$‍