1. \(sin60^031'=cos\left(90^0-60^031'\right)=cos29^029'\)

\(cos75^012'=sin\left(90-75^012'\right)=sin14^048'\)

\(cot80^0=\dfrac{1}{tan80^0}=tan\left(90^0-80^0\right)=tan10^0\)

\(tan57^030'=\dfrac{1}{tan\left(90^0-57^030'\right)}=\dfrac{1}{tan32^030'}\)

2.a)  \(BC=\sqrt{AB^2+AC^2}=\sqrt{21^2+72^2}=75\left(cm\right)\)

Ta có: \(sinB=\dfrac{AC}{BC}=\dfrac{72}{75}=\dfrac{24}{25}\Rightarrow\angle B\approx74\)

\(\Rightarrow\angle C\approx16\)

b) Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{21.72}{75}=\dfrac{504}{25}\left(cm\right)\)

Ta có: \(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=CH.BC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{21^2}{75}=\dfrac{147}{25}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{72^2}{75}=\dfrac{1728}{25}\left(cm\right)\end{matrix}\right.\)

c) Vì BD là phân giác góc B \(\Rightarrow\dfrac{AB}{BC}=\dfrac{AD}{CD}\)

\(\Rightarrow\dfrac{AD}{CD}=\dfrac{7}{25}\Rightarrow\dfrac{CD}{AD}=\dfrac{25}{7}\Rightarrow CD=\dfrac{25}{7}AD\)

Ta có: \(AD+CD=AC\Rightarrow AD+\dfrac{25}{7}AD=72\Rightarrow\dfrac{32}{7}AD=72\)

\(\Rightarrow AD=\dfrac{63}{4}\) (cm)

3. Kẻ đường cao BH

Ta có: \(BC^2=BH^2+HC^2=AB^2-AH^2+\left(AC-AH\right)^2\)

\(=AB^2-AH^2+AC^2+AH^2-2AC.AH=AB^2+AC^2-2.AC.AH\left(1\right)\)

Ta có: \(cosBAC=\dfrac{AH}{AB}\Rightarrow cos60=\dfrac{AH}{AB}\Rightarrow\dfrac{AH}{AB}=\dfrac{1}{2}\Rightarrow2AH=AB\left(2\right)\)

Thế (2) vào (1) \(\Rightarrow BC^2=AB^2+AC^2-AC.AB\)