登录
例 2 等腰三角形$ABC$中，若$BC = 3$，周长为$7$，求$\angle B$的余弦值.
答案
分析：构造直角三角形，将$\angle B$和$\angle C$分别放在直角三角形中，利用公共边构造方程.
解：分三种情况：
①当$BC$为底边时(如图①)，则$AB = AC = 2$，作$AD \perp BC$于点$D$，在$ Rt \bigtriangleup ABD$中，$BD = \frac{3}{2}$，$\therefore \cos B = \frac{BD}{AB} = \frac{3}{2} × \frac{1}{2} = \frac{3}{4}$.
②当$AC$为底边时(如图②)，则$AB = BC = 3$，$AC = 1$，作$AD \perp BC$于点$D$.
在$ Rt \bigtriangleup ABD$中，$AD^{2} = AB^{2} - BD^{2}$，在$ Rt \bigtriangleup ACD$中，$AD^{2} = AC^{2} - CD^{2}$，
$\therefore AC^{2} - CD^{2} = AB^{2} - BD^{2}$.
$\therefore BD^{2} - CD^{2} = AB^{2} - AC^{2} = 8$，$BD + CD = BC = 3$.
$\therefore BD = \frac{17}{6}$ $\therefore \cos B = \frac{BD}{AB} = \frac{17}{6} × \frac{1}{3} = \frac{17}{18}$.
③当$AB$为底边时(如图③)，作$AD \perp BC$于点$D$，则$AB = 1$，$BD = \frac{1}{6}$，
$\therefore \cos B = \frac{BD}{AB} = \frac{1}{6}$.
综上，$\cos B = \frac{3}{4}$或$\frac{17}{18}$或$\frac{1}{6}$.
