a: Xét ΔABD có \(cosBAD=\dfrac{AB^2+AD^2-BD^2}{2\cdot AB\cdot AD}\)
=>\(\dfrac{5^2+3^2-BD^2}{2\cdot5\cdot3}=cos120=-\dfrac{1}{2}\)
=>\(34-BD^2=-15\)
=>BD=7
\(S_{BAD}=\dfrac{1}{2}\cdot AB\cdot AD\cdot sinBAD=\dfrac{1}{2}\cdot5\cdot3\cdot sin120\)
\(=\dfrac{1}{2}\cdot15\cdot\dfrac{\sqrt{3}}{2}=\dfrac{15\sqrt{3}}{4}\)
ABCD là hình bình hành
=>\(S_{ABCD}=2\cdot S_{BAD}=2\cdot\dfrac{15\sqrt{3}}{4}=\dfrac{15\sqrt{3}}{2}\)
b: \(\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{DA}\)
\(=\overrightarrow{BC}+\overrightarrow{DA}\)
\(=\overrightarrow{AD}+\overrightarrow{DA}=\overrightarrow{0}\)
c: ABCD là hình bình hành
=>\(\overrightarrow{CB}=\overrightarrow{DA}\)
\(\overrightarrow{CB}+\overrightarrow{AD}+\overrightarrow{AC}=\overrightarrow{DA}+\overrightarrow{AD}+\overrightarrow{AC}=\overrightarrow{AC}\)
