by Cho tam giác ABC có $\widehat{A}={{50}^{0}}$
. Tìm tổng $\left( \overrightarrow{BA},\overrightarrow{CB} \right)+\left( \overrightarrow{CA},\overrightarrow{BC} \right)$ .
A. $ {{130}^{0}}. $
B. $ {{230}^{0}}. $
C. $ {{150}^{0}}. $
D. $ {{250}^{0}}. $
Hướng dẫn
Ta có: $ \widehat{A}+\widehat{B}+\widehat{C}={{180}^{0}}\Rightarrow \widehat{B}+\widehat{C}={{180}^{0}}-\widehat{A}={{130}^{0}}. $
$\left( \overrightarrow{BA},\overrightarrow{CB} \right)+\left( \overrightarrow{CA},\overrightarrow{BC} \right)={{180}^{0}}-\widehat{B}+{{180}^{0}}-\widehat{C}$
$={{360}^{0}}-\left( \widehat{B}+\widehat{C} \right)={{360}^{0}}-{{130}^{0}}={{230}^{0}}$Chọn đáp án B.
