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题目展示——先探究、再听讲
题目2
在锐角$\triangle ABC$中，角$A$，$B$，$C$所对的边为$a$，$b$，$c$，若$\displaystyle\frac{\sin B\sin C}{3\sin A} = \frac{\cos A}{a} + \frac{\cos C}{c}$且$\sin^2 A + \sin^2 B – \sin^2 C = \sin A \cdot \sin B$，求$\displaystyle\frac{c^2}{a + b}$的取值范围。解题探究——点击直达
答案解析
\(\sin^{2}A + \sin^{2}B – \sin^{2}C = \sin A\sin B\)\(\Rightarrow a^{2} + b^{2} – c^{2} = ab\)
\(\cos C = \displaystyle \frac{a^{2} + b^{2} – c^{2}}{2ab} = \displaystyle \frac{1}{2}\)
\(C \in (0, \pi) \Rightarrow C = \displaystyle \frac{\pi}{3}\)
\(\because \displaystyle \frac{\sin B\sin C}{3\sin A} = \displaystyle \frac{\cos A}{a} + \displaystyle \frac{\cos C}{c}\)
\(\therefore \displaystyle \frac{\displaystyle \frac{b}{2R}\sin C}{3\displaystyle \frac{a}{2R}} = \displaystyle \frac{c\cos A + a\cos C}{ac}\)
\(\displaystyle \frac{b\sin \displaystyle \frac{\pi}{3}}{3a} = \displaystyle \frac{c\cos A + a\cos C}{ac}\)
如图：\(A\)
\(c\cos A = AD\)，\(a\cos C = CD\)
\(\therefore c\cos A + a\cos C = AC = b\)
即原式为\(\displaystyle \frac{\displaystyle \frac{\sqrt{3}}{2}b}{3a} = \displaystyle \frac{b}{ac}\)
\(\Rightarrow \displaystyle \frac{\sqrt{3}}{6} = \displaystyle \frac{1}{c}\) \(\therefore c = 2\sqrt{3}\)
\(C\)为定角\(\displaystyle \frac{\pi}{3}\)
\(c\)为定弦\(2\sqrt{3} \Rightarrow\)外接圆
如图，\(\triangle ABC\)为等边\(\triangle\)时\((a + b)_{\max} = 4\sqrt{3}\)
\(BC\)为\(2R\)即\(\triangle ABC\)为\(Rt\triangle\)时\((a + b)_{\min} = 6\)
\(\therefore a + b \in (6, 4\sqrt{3}]\)
当\(a + b\)最大时\(\displaystyle \frac{c}{a + b}\)最小
\(\therefore \displaystyle \frac{c^{2}}{a + b} \in [\sqrt{3}, 2)\)