本题由2023级25班王雨婷讲解！
[ratemypost]
题目展示——先探究、再听讲
题目1
如图，在$\triangle ABC$中，$\angle BAC=\displaystyle\frac{2\pi}{3}$，点$D$在线段$BC$上，$AD\perp AC$，$\displaystyle\frac{BD}{CD}=\frac{1}{4}$，求$\sin C.$
解题探究——点击直达
答案解析
设\(BD = 1\)，\(CD = 4\)，\(b = 4\cos C\)\(\triangle ABD\)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\(\cos^{2}C = \displaystyle \frac{25}{28}\)，\(\therefore \sin^{2}C = \displaystyle \frac{3}{28}\)
\(C \in (0, \pi)\) \(\therefore \sin C = \displaystyle \frac{\sqrt{21}}{14}\)．