심화 문제: 내접한 도형

$\begin{array}{rl}\text{원주}& =2\pi r\\ & =2\pi (15)\\ & =30\pi \end{array}$‍

$\begin{array}{rl}\mathrm{\angle }APB+{38}^{\circ }+{38}^{\circ }& ={180}^{\circ }\\ \mathrm{\angle }APB& ={104}^{\circ }\end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{호의 길이}}{\text{원주}}}& ={\displaystyle \frac{\text{호의 중심각}}{\text{원의 중심각}}}\\ \\ \\ {\displaystyle \frac{\text{호의 길이}}{30\pi }}& ={\displaystyle \frac{{104}^{\circ }}{{360}^{\circ }}}\\ \\ \\ \text{호의 길이}& ={\displaystyle \frac{{104}^{\circ }}{{360}^{\circ }}}\cdot 30\pi \\ \\ \\ & ={\displaystyle \frac{26}{3}}\pi \end{array}$‍

$\begin{array}{rl}\text{원주}& =2\pi r\\ & =2\pi (5)\\ & =10\pi \end{array}$‍

$\begin{array}{rl}\mathrm{\angle }BAD+{\displaystyle \frac{9}{20}}\pi & =\pi \\ \\ \\ \mathrm{\angle }BAD& ={\displaystyle \frac{11}{20}}\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{호의 길이}}{\text{원주}}}& ={\displaystyle \frac{\text{호의 중심각}}{\text{원의 중심각}}}\\ \\ \\ {\displaystyle \frac{\text{호의 길이}}{10\pi }}& ={\displaystyle \frac{\left({\displaystyle \frac{11}{10}}\pi \right)}{2\pi }}\\ \\ \\ \text{호의 길이}& ={\displaystyle \frac{11\pi }{10}}\cdot {\displaystyle \frac{10\pi }{2\pi }}\\ \\ \\ & ={\displaystyle \frac{11}{2}}\pi \end{array}$‍

$\begin{array}{rl}\text{원주}& =2\pi r\\ & =2\pi (5)\\ & =10\pi \end{array}$‍

$\begin{array}{rl}{\displaystyle \frac{\text{호의 길이}}{\text{원주}}}& ={\displaystyle \frac{\text{호의 중심각}}{\text{원의 중심각}}}\\ \\ \\ {\displaystyle \frac{\pi }{10\pi }}& ={\displaystyle \frac{\text{호의 중심각}}{2\pi }}\\ \\ \\ \text{호의 중심각}& =2\pi \cdot {\displaystyle \frac{\pi }{10\pi }}\\ \\ \\ & ={\displaystyle \frac{1}{5}}\pi \end{array}$‍

$\begin{array}{rl}0.3\pi +0.1\pi +\mathrm{\angle }BEC& =\pi \\ \mathrm{\angle }BEC& =0.6\pi \end{array}$‍