Câu 4. Trong mặt phẳng tọa độ Oxy, cho tam giác ABC có A(2; 4); B(-1; 1); C(-8;2).
a) Có: $\overrightarrow{BA}$ = (3; 3) ; $\overrightarrow{CB}$ = (7;-1); $\overrightarrow{CA}$ = (10;2)
$\left| \overrightarrow{BA} \right|=\sqrt{{{3}^{2}}+{{3}^{2}}}=3\sqrt{2}$
$\left| \overrightarrow{CB} \right|=\sqrt{({{7}^{2}}+{{(-1)}^{2}}}=5\sqrt{2}$
$\left| \overrightarrow{CA} \right|=\sqrt{{{10}^{2}}+{{2}^{2}}}=2\sqrt{26}$
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$ \cos \widehat{ACB}=\cos \left( \overrightarrow{CA};\overrightarrow{CB} \right)=\frac{\overrightarrow{CA}.\overrightarrow{CB}}{\left| \overrightarrow{CA} \right|.\left| \overrightarrow{CB} \right|}=\frac{68}{2\sqrt{26}.5\sqrt{2}}\approx 0,943$
$ \widehat{ACB}=19,{{44}^{o}}$
Có: $\overrightarrow{BA}$ = (3; 3); $\overrightarrow{BC}$ =(-7;1)
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$ \cos \widehat{ABC}=\cos \left( \overrightarrow{BA};\overrightarrow{BC} \right)=\frac{\overrightarrow{BA}.\overrightarrow{BC}}{\left| \overrightarrow{BA} \right|.\left| \overrightarrow{BC} \right|}=\frac{-18}{3\sqrt{2}.5\sqrt{2}}=-0,6$
$ \widehat{ABC}=126,{{87}^{o}}$
Xét tam giác ABC có: $\widehat{ABC}+\widehat{BAC}+\widehat{ACB}={{180}^{o}}$ (ĐL tổng ba góc trong tam giác)
- $\Rightarrow \widehat{BAC}={{180}^{o}}-\left( \widehat{BAC}+\widehat{ACB} \right)={{33,69}^{o}}$
b) Chu vi của tam giác ABC là:
$3\sqrt{2}$ + $5\sqrt{2}$ + $2\sqrt{26}$ $\approx$ 21,5.
c)
${{S}_{ABC}}=\frac{1}{2}.AH.BC$
${{S}_{ABM}}=\frac{1}{2}.AH.BM$
$ {{S}_{ABC}}=2{{S}_{ABM}}$
$ \Leftrightarrow \frac{1}{2}.AH.BC=2\frac{1}{2}.AH.BM\Leftrightarrow BC=2BM$
$\Leftrightarrow BM=\frac{1}{2}BC$
hay M là trung điểm của BC.