Trong tam giác vuông BCH:
\(sinB=\dfrac{CH}{BC}\Rightarrow CH=BC.sinB=12.sin60^0=6\sqrt{3}\left(cm\right)\)
\(\widehat{BCH}=90^0-\widehat{B}=30^0\Rightarrow\widehat{ACH}=\widehat{C}-\widehat{BCH}=10^0\)
Trong tam giác vuông ACH:
\(cos\widehat{ACH}=\dfrac{CH}{AC}\Rightarrow AC=\dfrac{CH}{cos\widehat{ACH}}=\dfrac{6\sqrt{3}}{cos10^0}\approx10,23\left(cm\right)\)
b.
\(tan\widehat{ACH}=\dfrac{AH}{CH}\Rightarrow AH=CH.tan\widehat{ACH}=6\sqrt{3}.tan10^0\approx1,83\left(cm\right)\)
\(cos\widehat{B}=\dfrac{BH}{BC}\Rightarrow BH=BC.cos\widehat{B}=12.cos60^0=6\left(cm\right)\)
\(\Rightarrow AB=AH+BH=7,83\left(cm\right)\)
\(S_{ABC}=\dfrac{1}{2}CH.AB\approx40,7\left(cm^2\right)\)
