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直角三角形三角函数回顾

什么是基本三角形的比率？

$\mathrm{sin}\left(\mathrm{\angle }A\right)=$$\frac{\text{BC}}{\text{AB}}$
$\mathrm{cos}\left(\mathrm{\angle }A\right)=$$\frac{\text{AC}}{\text{AB}}$
$\mathrm{tan}\left(\mathrm{\angle }A\right)=$$\frac{\text{BC}}{\text{AC}}$

练习: 求出边长

$\begin{array}{rl}\mathrm{sin}\left(\mathrm{\angle }B\right)& =\frac{AC}{AB}\\ \\ \mathrm{sin}\left({40}^{\circ }\right)& =\frac{AC}{7}\phantom{\rule{1em}{0ex}}\mathrm{\angle }B={40}^{\circ },AB=7\\ \\ 7\cdot \mathrm{sin}\left({40}^{\circ }\right)& =AC\end{array}$

$AC=7\cdot \mathrm{sin}\left({40}^{\circ }\right)\approx 4.5$

$BC=$

练习集 2: 求解一个角

$\begin{array}{rl}\mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{AC}{AB}\\ \\ \mathrm{cos}\left(\mathrm{\angle }A\right)& =\frac{6}{8}\phantom{\rule{1em}{0ex}}AC=6,AB=8\\ \\ \mathrm{\angle }A& ={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\end{array}$

$\mathrm{\angle }A={\mathrm{cos}}^{-1}\left(\frac{6}{8}\right)\approx {41.41}^{\circ }$

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