\(\triangle ABD\)中，\(\displaystyle \frac{BD}{\sin \displaystyle \frac{\pi}{6}} = \displaystyle \frac{c}{\sin \angle ADB}\)

所以，\(\sin \angle ADB = \sin \angle ADC = \cos C\)

即\(\displaystyle \frac{1}{\displaystyle \frac{1}{2}} = \displaystyle \frac{c}{\cos C}\)，\(c = 2\cos C\)

\(\triangle ABC\)中，\(c^{2} = b^{2} + 5^{2} – 2 \times b \times 5 \times \cos C\)

\(= 16\cos^{2}C + 25 – 40\cos^{2}C = 25 – 24\cos^{2}C\)

所以\(25 – 24\cos^{2}C = 4\cos^{2}C\)

得\(\cos^{2}C = \displaystyle \frac{25}{28}\)，\(\therefore \sin^{2}C = \displaystyle \frac{3}{28}\)

\(C \in (0, \pi)\) \(\therefore \sin C = \displaystyle \frac{\sqrt{21}}{14}\)．