Question

3．在△ABC中，內(nèi)角A，B，C的對(duì)邊分別為a，b，c，O是△ABC外接圓的圓心，若$\sqrt{2}αcosB=\sqrt{2}c-b$，且$\frac{cosB}{sinC}\overrightarrow{AB}+\frac{cosC}{sinB}\overrightarrow{AC}=m\overrightarrow{AO}$，則m的值是（　�。�

• A． $\frac{{\sqrt{2}}}{4}$
• B． $\frac{{\sqrt{2}}}{2}$
• C． $\sqrt{2}$
• D． $2\sqrt{2}$

Answer 1

分析 由$\sqrt{2}αcosB=\sqrt{2}c-b$，得$\sqrt{2}cosAsinB-sinB=0$，即cosA=$\frac{\sqrt{2}}{2}$，得A=$\frac{π}{4}$．
由$\frac{cosB}{sinC}\overrightarrow{AB}+\frac{cosC}{sinB}\overrightarrow{AC}=m\overrightarrow{AO}$，得$\frac{cosB}{snC}{\overrightarrow{AB}}^{2}+\frac{cosC}{sinB}\overrightarrow{AC}•\overrightarrow{AB}=m\overrightarrow{AO}•\overrightarrow{AB}$，
⇒$cosB+cosAcosC=\frac{1}{2}msinC$
則m=2×$\frac{cosB+cosAcosC}{sinC}$=2×$\frac{-cos（A+C）+cosAcosC}{sinC}$=2×$\frac{sinAsinC}{sinC}=2sinA$．

解答 解：∵$\sqrt{2}αcosB=\sqrt{2}c-b$，∴$\sqrt{2}sinAcosB=\sqrt{2}sin（A+B）-sinB$
⇒$\sqrt{2}sinAcosB=\sqrt{2}sinAcosB+\sqrt{2}cosAsinB-sinB\\;\\;\\;\$
⇒$\sqrt{2}cosAsinB-sinB=0$，∴cosA=$\frac{\sqrt{2}}{2}$，得A=$\frac{π}{4}$．
∵O是△ABC外接圓的圓心，∴$\overrightarrow{AO}•\overrightarrow{AB}=\frac{1}{2}{\overrightarrow{AB}}^{2}=\frac{1}{2}{c}^{2}$
由$\frac{cosB}{sinC}\overrightarrow{AB}+\frac{cosC}{sinB}\overrightarrow{AC}=m\overrightarrow{AO}$，得$\frac{cosB}{snC}{\overrightarrow{AB}}^{2}+\frac{cosC}{sinB}\overrightarrow{AC}•\overrightarrow{AB}=m\overrightarrow{AO}•\overrightarrow{AB}$，
⇒$\frac{cosB}{sinC}{c}^{2}+\frac{cosC}{sinB}bccosA=m×\frac{1}{2}{c}^{2}$⇒$\frac{cosB}{sinC}c+\frac{cosC}{sinB}bcosA=\frac{1}{2}mc$
⇒$cosB+cosAcosC=\frac{1}{2}msinC$
∴m=2×$\frac{cosB+cosAcosC}{sinC}$=2×$\frac{-cos（A+C）+cosAcosC}{sinC}$
=2×$\frac{sinAsinC}{sinC}=2sinA$=$\sqrt{2}$．
故選：C

點(diǎn)評(píng) 本題綜合考查了三角形的外接圓的性質(zhì)、向量的三角形法則、數(shù)量積運(yùn)算、正弦定理、三角形的內(nèi)角和定理、兩角和的圓心公式等基礎(chǔ)知識(shí)與基本技能，考查了數(shù)形結(jié)合的能力、推理能力、計(jì)算能力．