Đáp án:
$V_{S.ABC} = \dfrac{a^3}{2}$
Giải thích các bước giải:
Trong $mp(ABC)$ kẻ $AH\perp BC$
Ta có:
$\begin{cases}AH\perp BC\quad \text{(cách dựng)}\\SA\perp BC\quad (SA\perp (ABC))\end{cases}$
$\Rightarrow BC\perp (SAH)$
$\Rightarrow BC\perp SH$
Khi đó:
$\begin{cases}(SBC)\cap (ABC) = BC\\SH\perp BC\quad (cmt)\\SH\subset (SBC)\\AH\perp BC\quad \text{(cách dựng)}\\AH\subset (ABC)\end{cases}$
$\Rightarrow \widehat{((SBC);(ABC))}= \widehat{SHA} = 60^\circ$
$\Rightarrow \begin{cases}SH = \dfrac{SA}{\sin\widehat{SHA}} = \dfrac{a}{\sin60^\circ} = \dfrac{2a\sqrt3}{3}\\AH = \dfrac{SA}{\tan\widehat{SHA}} = \dfrac{a}{\tan60^\circ} = \dfrac{a\sqrt3}{3}\end{cases}$
Mặt khác:
$BC\perp SH\quad (cmt)$
$\Rightarrow S_{SBC} = \dfrac12BC.SH$
$\Rightarrow BC = \dfrac{2S_{SBC}}{SH} = \dfrac{2\cdot 3a^2}{\dfrac{2a\sqrt3}{3}} = 3a\sqrt3$
$\Rightarrow S_{ABC} = \dfrac12BC.AH = \dfrac12\cdot 3a\sqrt3\cdot \dfrac{a\sqrt3}{3} = \dfrac{3a^2}{2}$
Ta được:
$\quad V_{S.ABC} = \dfrac13S_{ABC}.SA$
$\Leftrightarrow V_{S.ABC} = \dfrac13\cdot \dfrac{3a^2}{2}\cdot a = \dfrac{a^3}{2}$
