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Ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{A}=75^o\)
* \(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\Rightarrow AB=\dfrac{BCsinC}{sinA}=a\left(1+\sqrt{3}\right)\)
* \(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\Rightarrow AC=\dfrac{BCsinB}{sinA}=a\left(\dfrac{-6+3\sqrt{2}}{2}\right)\)
a) Có \(\overrightarrow{BC}^2=\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{AC}^2+\overrightarrow{AB}^2-2\overrightarrow{AC}.\overrightarrow{AB}\)
Suy ra: \(\overrightarrow{AC}.\overrightarrow{AB}=\dfrac{\overrightarrow{AC^2}+\overrightarrow{AB}^2-\overrightarrow{BC}^2}{2}=\dfrac{8^2+6^2-11^2}{2}=-\dfrac{21}{2}\).
Do \(\overrightarrow{AC}.\overrightarrow{AB}< 0\) nên \(cos\widehat{BAC}< 0\) suy ra góc A là góc tù.
b) Từ câu a suy ra: \(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=-\dfrac{21}{2.6.8}=-\dfrac{7}{32}\).
Do N là trung điểm của AC nên \(AN=AC:2=8:2=4cm\).
\(\overrightarrow{AM}.\overrightarrow{AN}=AM.AN.cos\left(\overrightarrow{AM},\overrightarrow{AN}\right)\)
\(=2.4.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=2.4.\dfrac{-7}{32}=-\dfrac{7}{4}\).
Gọi M là trung điểm của BC
Xét ΔABC có AM là đường trung tuyến
nên \(\overrightarrow{AB}+\overrightarrow{AC}=2\cdot\overrightarrow{AM}\)
\(\Leftrightarrow\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=2\cdot\dfrac{a\sqrt{3}}{2}=a\sqrt{3}\)
\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=a\)
A B C a
a) \(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=a.a.cos60^o=a.a.\dfrac{1}{2}\)\(=\dfrac{a^2}{2}\).
\(\overrightarrow{AB}.\overrightarrow{BC}=-\overrightarrow{BA}.\overrightarrow{BC}==-a.a.cos\left(\overrightarrow{BA},\overrightarrow{BC}\right)\)\(=-a.a.cos60^o=-\dfrac{a^2}{2}\).
\(\overrightarrow{AB}.\overrightarrow{AC}=\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=a.a.cos60=\dfrac{1}{2}a^2\)\(\overrightarrow{AB}.\overrightarrow{BC}=-\overrightarrow{BA}.\overrightarrow{BC}=-\left(\overrightarrow{BA}.\overrightarrow{BC}\right)=-\left(\left|\overrightarrow{BA}\right|.\left|\overrightarrow{BC}\right|.cos\left(\overrightarrow{BA},\overrightarrow{BC}\right)\right)=-\left(a.a.cos60\right)=-\dfrac{1}{2}a^2\)
góc C=180-75-45=60 độ
Xét ΔABC có AB/sinC=AC/sinB
=>AB/sin60=2/sin45
=>\(AB=\sqrt{6}\)
Ta có:
\(\widehat{C}=180^o-75^o-45^o=60^o\)
Xét tam giác ABC ta có:
\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)
\(\Rightarrow AB=\dfrac{ACsinC}{sinB}\)
\(\Rightarrow AB=\dfrac{2\cdot sin60^o}{sin45^o}\)
\(\Rightarrow AB=\sqrt{6}\)
Vậy: ...