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 $\sin A > 0$： $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$$ 2. **利用餘弦定理求對邊 $a = \overline{BC}$**： $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$a^2 = (4\sqrt{5})^2 + 8^2 - 2(4\sqrt{5})(8) \left(\frac{1}{\sqrt{5}}\right)$$ $$a^2 = 80 + 64 - 64 = 80 \implies a = \sqrt{80} = 4\sqrt{5}$$ 3. **利用正弦定理求 $\sin C$**： $$\frac{a}{\sin A} = \frac{c}{\sin C} \implies \frac{4\sqrt{5}}{2/\sqrt{5}} = \frac{8}{\sin C}$$ $$\frac{4\sqrt{5} \cdot \sqrt{5}}{2} = \frac{8}{\sin C} \implies 10 = \frac{8}{\sin C}$$ $$\sin C = \frac{8}{10} = \frac{4}{5}$$ 故所求 $\sin \angle ACB = \frac{4}{5}$。