15:
a: ΔABC cân tại A
=>góc ABC=góc ACB=(180-120)/2=30 độ
Xét ΔABC có
\(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)
=>BC/sin120=6/sin30
=>BC=6*căn 3(cm)
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC=\dfrac{1}{2}\cdot6\cdot6\cdot\dfrac{\sqrt{3}}{2}=\dfrac{18\sqrt{3}}{2}=9\sqrt{3}\left(cm^2\right)\)
b: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)
\(\overrightarrow{MB}-\overrightarrow{MC}=\overrightarrow{CM}+\overrightarrow{MB}=\overrightarrow{CB}\)
Do đó: \(\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{MB}-\overrightarrow{MC}\)
=>\(\overrightarrow{AB}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{AC}\)
c: \(\left|\overrightarrow{AB}-\overrightarrow{CM}\right|=\left|\overrightarrow{AB}+\overrightarrow{MC}\right|\)
\(=\left|\overrightarrow{AB}+\dfrac{1}{2}\left(\overrightarrow{BC}\right)\right|=\left|\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AC}\right|\)
\(=\dfrac{1}{2}\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\dfrac{1}{2}\cdot\left|\left(2\cdot\overrightarrow{AM}\right)\right|=\left|\overrightarrow{AM}\right|=AM\)
ΔAMB vuông tại M nên AB^2=AM^2+BM^2
=>AM^2=6^2-(3căn 3)^2=9
=>AM=3(cm)
=>\(\left|\overrightarrow{AB}-\overrightarrow{CM}\right|=3\left(cm\right)\)