Cho khối chóp $S.ABC$ có đáy $ABC$ là tam giác cân đỉnh $A$, góc...

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+ Gọi $HS\left( ABC \right)SA=SB=SCHABCSA\left( ABC \right)\angle SAH\Rightarrow \angle SAH={{60}^{0}}B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2.AB.AC.cos\left( \angle BAC \right)=3{{a}^{2}}\Rightarrow BC=a\sqrt{3}.\frac{BC}{\sin \angle BAC}=2AH\Rightarrow AH=aSH=AH.\tan \angle SAH=a\sqrt{3}{{V}_{S.ABC}}=\frac{1}{3}.{{S}_{\Delta ABC}}.SH=\frac{1}{6}.AB.AC.sin{{120}^{0}}.SH=\frac{{{a}^{3}}}{4}$.