\(\dfrac{a}{SinA}=\dfrac{b}{SinB}=\dfrac{c}{SinC}\)
\(\Rightarrow\dfrac{4}{Sin105}=\dfrac{AC}{Sin45}=\dfrac{AB}{Sin30}=4\sqrt{6}-4\sqrt{2}\)
\(\Rightarrow\left\{{}\begin{matrix}AB=2\sqrt{6}-2\sqrt{2}\\AC=-4+4\sqrt{3}\end{matrix}\right.\) ( cm )
Vậy ..
Vẽ đường cao AH
Ta có: ΔAHB vuông tại H(gt)
mà \(\widehat{B}=45^0\)
nên ΔAHB vuông cân tại H
⇔\(BH=AH=HC\cdot\tan30^0\)
\(\Leftrightarrow BH-CH\cdot\tan30^0=\dfrac{CH}{\sqrt{3}}\)
\(\Leftrightarrow BH=\dfrac{4}{1+\sqrt{3}}\)
\(\Leftrightarrow AB=\dfrac{AH}{\sin45^0}\simeq2,06\left(cm\right)\)
\(\Leftrightarrow AC=2\cdot AH=2,92\left(cm\right)\)
