浙江 2015 · 数学 q16

题面

16．（14 分）（2015•浙江）在$\triangle \mathrm{A B C}$中，内角A，B，C所对的边分别为a，b，c，已知\mathrm{A}{=} \frac{\pi}{\updownarrow} ;$$b^{2} - a^{2} = \frac{1}{2} c^{2}

（1）求tanC的值；

（2）若$\triangle \mathrm{A B C}$的面积为3，求b的值

解析

分析

（1）由余弦定理可得：$a^{2} = b^{2} + c^{2} - 2 b c \cos \frac{\pi}{4}$，已知$b^{2} - a^{2} = \frac{1}{2} c^{2}$．可得b = \frac{3 \sqrt{2} c}{4} ,$$\mathrm{a}{=} \frac{\sqrt{1 0}}{4} c .．利用余弦定理可得cosC．可得$\mathrm{s i n C =}{\sqrt{1 - \cos^{2} \complement}}$，即可得出$\mathrm{t a n C}{=} \frac{s i n C}{c o s C} .$（2）由$S_{\Delta A B C} = \frac{1}{2} a b s i n c = \frac{1}{2} \times \frac{\sqrt{1 0}}{4} c \times \frac{3 \sqrt{2}}{4} c \times \frac{2 \sqrt{5}}{5} - 3$，可得c，即可得出b解答：解：（1）$\because A = {\frac{\pi}{4}} ;$，∴由余弦定理可得：$a^{2} = b^{2} + c^{2} - 2 b c \cos \frac{\pi}{4} , \therefore b^{2} - a^{2} = \sqrt{2} b c - c^{2}$又$b^{2} - a^{2} = \frac{1}{2} c^{2} . \therefore \sqrt{2} 0 c - c^{2} = \frac{1}{2} c^{2} . \therefore \sqrt{2} b = \frac{3}{2} c$．可得b = \frac{3 \sqrt{2} c}{4} ,$$\therefore a^{2} = b^{2} - \frac{1}{2} c^{2} = \frac{5}{8} c^{2}，即\mathrm{a}{=} \frac{\sqrt{1 0}}{4} \varsigma$$\therefore \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} - \frac{c^{2}}{8} c^{2} + \frac{9}{8} c^{2} - c^{2}{2 \times \frac{\sqrt{1 0}}{4} c \times \frac{3 \sqrt{2}}{4} c} - \frac{\sqrt{5}}{5} .$$\because{\mathrm{C}} \in \ ( 0 , \ \pi )$$\therefore \sin C = \sqrt{1 - \cos^{2} C} = \frac{2 \sqrt{5}}{5} .$$\therefore \mathrm{t a n C} = \frac{s i n C}{c o s C} = 2 .（2）$\because S_{\Delta \Delta B C} = \frac{1}{2} \pi \Delta \sin C = \frac{1}{2} \times \frac{\sqrt{1 0}}{4} \times \frac{3 \sqrt{2}}{4} \times \frac{2 \sqrt{5}}{5} = 3 .$解得$\mathrm{{c} =} 2 \sqrt{2} .$

$b = \frac{3 \sqrt{2} c}{4} = 3 .$