Theo giả thiết: $\angle SBA=\alpha$ .Ta có: $AC=AB\Rightarrow SB=SC \Rightarrow SD\perp BC$ Mà $AD\perp BC \Rightarrow BC\perp(SAD) \Rightarrow \angle BSD=\beta$.Ta có: $AB=SB.\cos\alpha, BD=SB.\sin\beta$$\Rightarrow a^2=AD^2=SB^2(\cos^2\alpha-\sin^2\beta) \Rightarrow SB=\frac{a}{\sqrt{ \cos^2\alpha-\sin^2\beta }}$ Suy ra: $BD=SB\sin\beta= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ $SA=SB\sin\alpha= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ Từ đó: $V_{S.ABC}= \frac{a^3\sin\alpha\sin\beta}{3(\cos^2\alpha-\sin^2\beta)} $
Theo giả thiết: $\angle SBA=\alpha$ .Ta có: $AC=AB\Rightarrow SB=SC \Rightarrow SD\perp BC$ Mà $AD\perp BC \Rightarrow BC\perp(SAD) \Rightarrow BSD=\beta$.Ta có: $AB=SB.\cos\alpha, BD=SB.\sin\beta$$\Rightarrow a^2=AD^2=SB^2(\cos^2\alpha-\sin^2\beta) \Rightarrow SB=\frac{a}{\sqrt{ \cos^2\alpha-\sin^2\beta }}$ Suy ra: $BD=SB\sin\beta= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ $SA=SB\sin\alpha= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ Từ đó: $V_{S.ABC}= \frac{a^3\sin\alpha\sin\beta}{3\sqrt{ \cos^2\alpha-\sin^2\beta }} $
Theo giả thiết: $\angle SBA=\alpha$ .Ta có: $AC=AB\Rightarrow SB=SC \Rightarrow SD\perp BC$ Mà $AD\perp BC \Rightarrow BC\perp(SAD) \Rightarrow
\angle BSD=\beta$.Ta có: $AB=SB.\cos\alpha, BD=SB.\sin\beta$$\Rightarrow a^2=AD^2=SB^2(\cos^2\alpha-\sin^2\beta) \Rightarrow SB=\frac{a}{\sqrt{ \cos^2\alpha-\sin^2\beta }}$ Suy ra: $BD=SB\sin\beta= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ $SA=SB\sin\alpha= \frac{a\sin\beta}{\sqrt{ \cos^2\alpha-\sin^2\beta }} $ Từ đó: $V_{S.ABC}= \frac{a^3\sin\alpha\sin\beta}{3
(\cos^2\alpha-\sin^2\beta
)} $