<選填>設銳角三角形 \(ABC\) 的外接圓半徑為 8。已知外接圓圓心到 \(AB\) 的距離為 2,而到 \(BC\) 的距離為 7,則 \(AC =\)__________。
設 \(\angle OBM = \theta_1\)，\(\angle OBN = \theta_2\)
\[
\Rightarrow BM = \sqrt{BO^2 - OM^2} = \sqrt{8^2 - 2^2} = \sqrt{60} = 2\sqrt{15}
\]
\[
BN = \sqrt{BO^2 - ON^2} = \sqrt{8^2 - 7^2} = \sqrt{15}
\]
\[
\therefore \cos\theta_1 = \frac{BM}{BO} = \frac{2\sqrt{15}}{8} = \frac{\sqrt{15}}{4},\quad \sin\theta_1 = \frac{OM}{BO} = \frac{2}{8} = \frac{1}{4}
\]
\[
\cos\theta_2 = \frac{BN}{BO} = \frac{\sqrt{15}}{8},\quad \sin\theta_2 = \frac{ON}{BO} = \frac{7}{8}
\]
\[
\Rightarrow \sin(\angle ABC) = \sin(\theta_1 + \theta_2) = \sin\theta_1 \cos\theta_2 + \cos\theta_1 \sin\theta_2
\]
\[
= \frac{1}{4} \times \frac{\sqrt{15}}{8} + \frac{\sqrt{15}}{4} \times \frac{7}{8} = \frac{\sqrt{15}}{32} + \frac{7\sqrt{15}}{32} = \frac{8\sqrt{15}}{32} = \frac{\sqrt{15}}{4}
\]
\[
\therefore \text{由正弦定理得 } \frac{AC}{\sin(\angle ABC)} = 2R \Rightarrow AC = 2R \cdot \sin(\angle ABC) = 2 \times 8 \times \frac{\sqrt{15}}{4} = 4\sqrt{15}
\]
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