Lời giải:
a) Xét tam giác vuông $AHB$ vuông tại $H$ ta có:
\(\tan \widehat{ABH}=\frac{AH}{HB}\Leftrightarrow \frac{\sqrt{3}}{3}=\tan 30^0=\frac{AH}{BH}\)
\(\Leftrightarrow AH=\frac{\sqrt{3}BH}{3}=2\sqrt{3}\) (cm)
Xét tam giác $ACH$ vuông tại $H$ ta có:
\(\sin \widehat{ACH}=\frac{AH}{AC}\Leftrightarrow AC=\frac{AH}{\sin 50^0}=\frac{2\sqrt{3}}{\sin 50^0}\) (cm)
b)
Ta có: \(\tan \widehat{ACH}=\frac{AH}{CH}\Leftrightarrow CH=\frac{AH}{\tan \widehat{ACH}}=\frac{2\sqrt{3}}{\tan 50^0}\) (cm)
\(S_{ACH}=\frac{AH.CH}{2}=\frac{2\sqrt{3}.2\sqrt{3}}{2\tan 50}=\frac{6}{\tan 50}\) (cm2 )
\(C_{ACH}=AC+CH+AH=\frac{2\sqrt{3}}{\sin 50}+\frac{2\sqrt{3}}{\tan 50}+2\sqrt{3}\approx 10,9\) (cm)
