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PART 13：正弦定律-例題
於 \(\vartriangle ABC\) ，若 \(\overline {BC} = 10\) ， \(\angle BAC = \frac{\pi }{6}\) ， \(\angle BCA = \frac{\pi }{4}\) ，
則 \(\overline {AB} = ?\)
圖18. \(\vartriangle ABC\) 邊角關係 |
\(\frac{{10}}{{\sin \frac{\pi }{6}}} = \frac{{\overline {AB} }}{{\sin \frac{\pi }{4}}}\) \(\quad \Rightarrow \quad \frac{{10}}{{\frac{1}{2}}} = \frac{{\overline {AB} }}{{\frac{{\sqrt 2 }}{2}}}\)
\( \Rightarrow \quad 20 \cdot \frac{{\sqrt 2 }}{2} = \overline {AB} \)\(\quad \Rightarrow \quad \overline {AB} = 10\sqrt 2 \)
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