Gọi H là trung điểm BC
\(\left\{{}\begin{matrix}\left(SBC\right)\perp\left(ABC\right)\\\left(SBC\right)\cap\left(ABC\right)=BC\\SH\perp BC\end{matrix}\right.\Rightarrow SH\perp\left(ABC\right)\)
\(\left\{{}\begin{matrix}\widehat{\left[\left(SAB\right);\left(ABC\right)\right]=60^O}\\\left(SAB\right)\cap\left(ABC\right)=AB\\HN\perp AB\end{matrix}\right.\Rightarrow\widehat{SHN}=60^O\)
\(SH=\sqrt{SB^2-BH^2}=\sqrt{\left(\frac{\sqrt{3}a}{2}\right)^2-\left(\frac{a}{2}\right)^2}=\frac{a\sqrt{2}}{2}\)
\(HN=cot60.SH=\frac{\sqrt{3}}{3}.a\frac{\sqrt{2}}{2}=\frac{a\sqrt{6}}{6}\)
Xét tam giácABC ta có:
H là trung điểm BC
HN//AC
Suy ra HN là đường trung bình của tam giác ABC
\(\Rightarrow AC=2HN=\frac{a\sqrt{6}}{3}\)
\(AB=\sqrt{BC^2-AC^2}=\sqrt{a^2-\frac{2}{3}a^2}=\frac{a\sqrt{3}}{3}\)
\(V=\frac{1}{3}SH.S_{ABC}=\frac{1}{3}SH.\frac{1}{2}AB.AC=\frac{1}{6}.a\frac{\sqrt{2}}{2}.a\frac{\sqrt{6}}{6}.a\frac{\sqrt{6}}{3}=\frac{a^3\sqrt{2}}{36}\)
