Xét trong [imath]\Delta ABC[/imath] vuông tại [imath]B[/imath] có: [imath]\widehat{A}+\widehat{C}=90\degree \Rightarrow \widehat{C}=50\degree[/imath]
Ta có:
+ [imath]\displaystyle \cos{A}=\frac{AB}{AC} \Leftrightarrow AC=\frac{AB}{cos{A}}=\frac{2\sqrt{3}.a}{3}[/imath]
+ [imath]\displaystyle \tan{A}=\frac{BC}{BA} \Leftrightarrow BC=BA.\tan{A}=\frac{a\sqrt{3}}{3}[/imath]
a) [imath]\left| \overrightarrow{BA}+2.\overrightarrow{BC} \right|=\left|\overrightarrow{IA}-\overrightarrow{IB}+2.\left(\overrightarrow{IC}-\overrightarrow{IB} \right) \right|=\left|\overrightarrow{IA}+2.\overrightarrow{IC}\right|=\left| 0+\overrightarrow{IC}\right| ~ \left( 1 \right)[/imath]
[imath]\displaystyle \Rightarrow IC=\frac{AC}{2}=\frac{a\sqrt{3}}{3}[/imath]
b) [imath]\left| \overrightarrow{AB}-\overrightarrow{AC}+3.\overrightarrow{IA}\right|=\left| \overrightarrow{IB}-\overrightarrow{IA}-\left( \overrightarrow{IC}-\overrightarrow{IA} \right)+3.\overrightarrow{IA} \right|=\left| \left( \overrightarrow{IB}-\overrightarrow{IC} \right)+3.\overrightarrow{IA}\right|=\left|\overrightarrow{CB}+3.\overrightarrow{IA}\right|~\left( 2 \right)[/imath]
Bình phương (2) lên ta được:
[imath]CB^2+9IA^2+6.\overrightarrow{CB}.\overrightarrow{IA}=CB^2+9.IA^2+6.\left| \overrightarrow{CB} \right|.\left| \overrightarrow{IA} \right|. \cos{\widehat{IAB}} \Rightarrow \sqrt{CB^2+9.IA^2+6.\left| \overrightarrow{CB} \right|.\left| \overrightarrow{IA} \right|. \cos{\widehat{IAB}}}[/imath]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
