BT1:
\(\dfrac{m}{n}=\dfrac{e}{f}\Rightarrow\left\{{}\begin{matrix}m=\dfrac{e\cdot n}{f}\\n=\dfrac{m\cdot f}{e}\\e=\dfrac{m\cdot f}{n}\\f=\dfrac{e\cdot n}{m}\end{matrix}\right.\)
BT2:
\(BC=\dfrac{AB}{\sin C}=\dfrac{11}{\sin60^0}=11\cdot\dfrac{2}{\sqrt{3}}=\dfrac{22\sqrt{3}}{3}\left(cm\right)\\ AC=\sqrt{BC^2-AB^2}=\dfrac{11\sqrt{3}}{3}\left(cm\right)\left(pytago\right)\\ \widehat{B}=90^0-\widehat{C}=30^0\)
BT4:
\(AH=\sin B\cdot AB=\sin30^0\cdot12=\dfrac{1}{2}\cdot12=6\left(cm\right)\\ HB=\sqrt{AB^2-AH^2}=6\sqrt{3}\left(cm\right)\left(pytago\right)\)
Vì tg ABC cân tại A nên AH cx là trung tuyến
Do đó \(HB=HC=6\sqrt{3}\left(cm\right)\)
