Giải thích các bước giải:

a.Kẻ $BD\perp AC$ 

Ta có:

$\sin C=\dfrac{BD}{CB}$

$\to BD=BC\sin C$

$\to BD=20\sin30^o$

$\to DB=10$

b.Ta có:

$\hat A=180^o-\hat B-\hat C= 128^o$

$\widehat{DAB}=180^o-\widehat{BAC}=52^o$

$\sin\widehat{DAB}=\dfrac{BD}{AB}\to AB=\dfrac{BD}{\sin\widehat{DAB}}=\dfrac{10}{\sin52^o}\approx 12.7$

$CD=\sqrt{BC^2-DB^2}=\sqrt{20^2-10^2}=10\sqrt3$

$AD=\sqrt{AB^2-BD^2}=\sqrt{12.7^2-10^2}=7.8$

$AC=CD-AD=CD-AC=10\sqrt3-7.8\approx 9.5$

c.Kẻ $AE\perp BC$

$\to AE.BC=DB.AC(=2S_{ABC})$

$\to AE=\dfrac{BD.AC}{BC}=\dfrac{10\cdot 9.5}{20}=4.75$