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# 正弦和余弦定理回顾

## 正弦定理

$\frac{a}{\mathrm{sin}\left(\alpha \right)}=\frac{b}{\mathrm{sin}\left(\beta \right)}=\frac{c}{\mathrm{sin}\left(\gamma \right)}$

## 余弦定理

${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\left(\gamma \right)$

## 练习集 1: 使用正弦定理求解三角形

### 例子一: 求出缺失的边

$\begin{array}{rl}\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}& =\frac{AC}{\mathrm{sin}\left(\mathrm{\angle }B\right)}\\ \\ \frac{5}{\mathrm{sin}\left({33}^{\circ }\right)}& =\frac{AC}{\mathrm{sin}\left({67}^{\circ }\right)}\\ \\ \frac{5\mathrm{sin}\left({67}^{\circ }\right)}{\mathrm{sin}\left({33}^{\circ }\right)}& =AC\\ \\ 8.45& \approx AC\end{array}$

### 例子二: 求出缺失的角度

$\begin{array}{rl}\frac{BC}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}\\ \\ \frac{11}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{5}{\mathrm{sin}\left({25}^{\circ }\right)}\\ \\ 11\mathrm{sin}\left({25}^{\circ }\right)& =5\mathrm{sin}\left(\mathrm{\angle }A\right)\\ \\ \frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}& =\mathrm{sin}\left(\mathrm{\angle }A\right)\end{array}$

$m\mathrm{\angle }A={\mathrm{sin}}^{-1}\left(\frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}\right)\approx {68.4}^{\circ }$

$BC=$

## 练习集 2: 使用余弦定理求解三角形

### 例子一: 求出一个角

$\left(AC{\right)}^{2}=\left(AB{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AB\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }B\right)$

$\begin{array}{rl}\left(5{\right)}^{2}& =\left(10{\right)}^{2}+\left(6{\right)}^{2}-2\left(10\right)\left(6\right)\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 25& =100+36-120\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 120\mathrm{cos}\left(\mathrm{\angle }B\right)& =111\\ \\ \mathrm{cos}\left(\mathrm{\angle }B\right)& =\frac{111}{120}\end{array}$

$m\mathrm{\angle }B={\mathrm{cos}}^{-1}\left(\frac{111}{120}\right)\approx {22.33}^{\circ }$

### 例子二: 求出未知的边

$\left(AB{\right)}^{2}=\left(AC{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AC\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }C\right)$

$\begin{array}{rl}\left(AB{\right)}^{2}& =\left(5{\right)}^{2}+\left(16{\right)}^{2}-2\left(5\right)\left(16\right)\mathrm{cos}\left({61}^{\circ }\right)\\ \\ \left(AB{\right)}^{2}& =25+256-160\mathrm{cos}\left({61}^{\circ }\right)\\ \\ AB& =\sqrt{281-160\mathrm{cos}\left({61}^{\circ }\right)}\\ \\ AB& \approx 14.3\end{array}$

$m\mathrm{\angle }A=$
${}^{\circ }$

## 练习集 3: 普通三角形问题

"只剩余一个" 莱恩在他的藏身之处向他的兄弟发出信号.

"$34$ 度." 马特回复到,并告知莱恩和机器人之间的角度.