97指考數學乙選填第10題詳解

設 $\angle A = \angle BAC$、$\angle C = \angle ACB$。 已知 $c = \overline{AB} = 8$，$b = \overline{AC} = 4\sqrt{5}$，且 $\cos A = \frac{1}{\sqrt{5}}$。 1. 計算 $\sin A$： 因為 $0 < A < 180^\circ$，所以 $\sin A > 0$： $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\dfrac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \dfrac{1}{5}} = \sqrt{\dfrac{4}{5}} = \dfrac{2}{\sqrt{5}}$$ 2. 利用餘弦定理求對邊 $a = \overline{BC}$： \begin{aligned} a^2 &= b^2 + c^2 - 2bc \cos A\\[4pt] &= (4\sqrt{5})^2 + 8^2 - 2(4\sqrt{5})(8)\left(\dfrac{1}{\sqrt{5}}\right)\\[4pt] &= 80 + 64 - 64 = 80 \end{aligned} 所以 $a = \sqrt{80} = 4\sqrt{5}$。 3. 利用正弦定理求 $\sin C$： \begin{aligned} \dfrac{a}{\sin A} = \dfrac{c}{\sin C} \implies \dfrac{4\sqrt{5}}{2/\sqrt{5}} = \dfrac{8}{\sin C}\\[4pt] \dfrac{4\sqrt{5} \cdot \sqrt{5}}{2} = \dfrac{8}{\sin C} \implies 10 = \dfrac{8}{\sin C}\\[4pt] \sin C = \dfrac{8}{10} = \dfrac{4}{5} \end{aligned} 故所求 $\sin \angle ACB = \frac{4}{5}$。