Problemas de desafío: longitud de arco 2

$\begin{array}{rl}\text{Circunferencia}& =2\pi r\\ & =2\pi (10)\\ & =20\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{longitud del arco}}{\text{circunferencia}}}& ={\displaystyle \frac{\text{medida del arco}}{\text{grados en un círculo}}}\\ \\ \\ {\displaystyle \frac{16\pi }{20\pi }}& ={\displaystyle \frac{\text{medida del arco}}{{360}^{\circ }}}\\ \\ \\ \text{medida del arco}& ={360}^{\circ }\cdot {\displaystyle \frac{16\pi }{20\pi }}\\ \\ \\ & ={288}^{\circ }\end{array}$‍

$\begin{array}{r}{288}^{\circ }+\text{m}\stackrel{⌢}{AC}={360}^{\circ }\\ \text{m}\stackrel{⌢}{AC}={72}^{\circ }\end{array}$‍

$\begin{array}{rl}\text{Circunferencia}& =2\pi r\\ & =2\pi (18)\\ & =36\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{longitud del arco}}{\text{circunferencia}}}& ={\displaystyle \frac{\text{medida del arco}}{\text{grados en un círculo}}}\\ \\ \\ {\displaystyle \frac{14\pi }{36\pi }}& ={\displaystyle \frac{\text{medida del arco}}{{360}^{\circ }}}\\ \\ \\ \text{medida del arco}& ={\displaystyle \frac{14\pi }{36\pi }}\cdot {360}^{\circ }\\ \\ \\ & ={140}^{\circ }\end{array}$‍

$\text{m}\stackrel{⌢}{BC}+{76}^{\circ }={140}^{\circ }$‍