giúp mình với

a,Áp dụng hệ thức lượng trong $\displaystyle \Delta BHC$
$\displaystyle \Rightarrow CH=BC.\sin\hat{B} =6.\sin 60^{o} =6.\frac{\sqrt{3}}{2} =3\sqrt{3} \ ( cm)$
Có $\displaystyle \hat{A} =180^{o} -60^{o} -40^{o} =80^{o}$
$\displaystyle \Rightarrow AC=\frac{CH}{\sin\hat{A}} =\frac{3\sqrt{3}}{\sin 80^{o}} \approx 5,3\ ( cm)$
b,Áp dụng định lí Pytago có
$\displaystyle HB=\sqrt{BC^{2} -CH^{2}} =\sqrt{6^{2} -\left( 3\sqrt{3}\right)^{2}} =\sqrt{36-27} =\ 3\ ( cm)$
$\displaystyle \Rightarrow AH=\sqrt{AC^{2} -CH^{2}} =\sqrt{( 5,3)^{2} -\left( 3\sqrt{3}\right)^{2}} \approx \sqrt{1,09\ } \ ( cm)$
$\displaystyle \Rightarrow AB=3+\sqrt{1,09} =\frac{30+\sqrt{109}}{10} \ ( cm)$
$\displaystyle \Rightarrow S_{ABC} =\frac{1}{2} .AB.CH=\frac{1}{2} .\frac{30+\sqrt{109}}{10} .3\sqrt{3} =\frac{3\sqrt{327} +90\sqrt{3}}{20} \ \ \left( cm^{2}\right)$