$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{a.Do\ tứ\ diện\ ABCD\ đều\ có\ cạnh\ là\ a.\Rightarrow \Delta ABC\ là\ tam\ giác\ đều}\\ \mathrm{\Rightarrow (\overrightarrow{AB} ,\overrightarrow{BC}) =\widehat{CBA} =60^{o}}\\ \mathrm{b.\Delta ABC\ là\ tam\ giác\ đều,\ CH\ là\ đường\ trung\ tuyến\Rightarrow CH\ cũng\ là\ đường\ cao}\\ \mathrm{\Rightarrow CH=\sqrt{AC^{2} -AH^{2}} =\frac{\sqrt{3}}{2} a}\\ \mathrm{Có\ sin\ \widehat{HCA} =\frac{AH}{AC} =\frac{1}{2} \Rightarrow \ \widehat{HCA} =30^{0}}\\ \mathrm{\Rightarrow (\overrightarrow{AC} ,\overrightarrow{CH}) =180^{o} -\widehat{HCA} =150^{o}}\\ \mathrm{c.(\overrightarrow{AD} ,\overrightarrow{BD}) =180^{o} -\widehat{ADB} =120^{o}}\\ \mathrm{d.\ }\\ \mathrm{Ta\ có:\overrightarrow{AD} .\overrightarrow{CH} =\overrightarrow{AD}\left(\frac{\mathrm{\overrightarrow{CA}} +\mathrm{\overrightarrow{CB}}}{2}\right) =\frac{1}{2}(\overrightarrow{AD} .\overrightarrow{CA} +\ \overrightarrow{AD} .\overrightarrow{CA} +\overrightarrow{AD} .\overrightarrow{AB})}\\ \mathrm{=\frac{1}{2}\left( a.a.cos120^{o} +a.a.cos120^{o} +a.a.cos60^{o}\right) =-\frac{1}{4} a^{2}}\\ \mathrm{Ta\ có:cos(\overrightarrow{AD} ,\overrightarrow{CH}) =\frac{\mathrm{\overrightarrow{AD} .\overrightarrow{CH}}}{|\mathrm{\overrightarrow{AD} |.|\overrightarrow{CH}} |} =\frac{-\frac{1}{4} a^{2}}{a.\frac{\sqrt{3}}{2} a} =-\frac{\sqrt{3}}{6}}\\ \mathrm{\Rightarrow (\overrightarrow{AD} ,\overrightarrow{CH}) \approx 106^{o} 47'} \end{array}$
