a) Theo giả thiết ta có$\begin{cases}CH \perp SH \\ CH \perp SA ( \text{do} SA \perp mp(ABC)) \end{cases}\Rightarrow CH \perp mp(SAH) \Rightarrow CH \perp AH$Do đó $HC=AC \cos \alpha =a \cos \alpha$Suy ra $S_{AHC}=\frac{1}{2}AC.HC.\sin \alpha=\frac{1}{4}a^2\sin 2\alpha$Suy ra $V_{S.AHC}=\frac{1}{3}SA.S_{AHC}=\frac{1}{12}a^3\sin 2\alpha \le \frac{1}{12}a^3.$Vậy $\max V_{S.AHC}= \frac{1}{12}a^3\Leftrightarrow \sin 2\alpha=1\Leftrightarrow \alpha=\frac{\pi}{4}$.
a) Theo giả thiết ta có$\begin{cases}CH \perp SH \\ CH \perp SA ( \text{do} SA \perp mp(ABC)) \end{cases}\Rightarrow CH \perp mp(SAH) \Rightarrow CH \perp AH$Do đó $HC=AC \cos \alpha =a \cos \alpha$Suy ra $S_{AHC}=\frac{1}{2}AC.HC.\sin \alpha=\frac{1}{4}a^2\sin 2\alpha$Suy ra $V_{S.AHC}=\frac{1}{3}SA.S_{AHC}=\frac{1}{12}a^3\sin 2\alpha \le \frac{1}{12}a^3.$Vậy $\min V_{S.AHC}= \frac{1}{12}a^3\Leftrightarrow \sin 2\alpha=1\Leftrightarrow \alpha=\frac{\pi}{4}$.
a) Theo giả thiết ta có$\begin{cases}CH \perp SH \\ CH \perp SA ( \text{do} SA \perp mp(ABC)) \end{cases}\Rightarrow CH \perp mp(SAH) \Rightarrow CH \perp AH$Do đó $HC=AC \cos \alpha =a \cos \alpha$Suy ra $S_{AHC}=\frac{1}{2}AC.HC.\sin \alpha=\frac{1}{4}a^2\sin 2\alpha$Suy ra $V_{S.AHC}=\frac{1}{3}SA.S_{AHC}=\frac{1}{12}a^3\sin 2\alpha \le \frac{1}{12}a^3.$Vậy $\m
ax V_{S.AHC}= \frac{1}{12}a^3\Leftrightarrow \sin 2\alpha=1\Leftrightarrow \alpha=\frac{\pi}{4}$.