Đáp án:

$\widehat{((SBC);(SCD))}=\arccos\dfrac13$

Giải thích các bước giải:

Ta có:

$\quad \begin{cases}AB\perp BC\\SH\perp BC\end{cases}$

$\Rightarrow BC\perp (SAB)$

Trong $mp(SAB)$ kẻ $HI\perp SB$

$\Rightarrow BC\perp HI$

$\Rightarrow HI\perp (SBC)$

Gọi $M$ là trung điểm $CD$

$\Rightarrow HM//BC//AD$

$\Rightarrow HM\perp CD;\ HM = HB = HA = BC = AD = a$

Ta được:

$\quad \begin{cases}HM\perp CD\\SH\perp CD\end{cases}$

$\Rightarrow CD\perp (SHM)$

Trong $mp(SHM)$ kẻ $HK\perp SM$

$\Rightarrow CD\perp HK$

$\Rightarrow HK\perp (SCD)$

Khi đó:

$\quad \begin{cases}HI\perp (SBC)\quad (cmt)\\HK\perp (SCD)\quad (cmt)\end{cases}$

$\Rightarrow \widehat{((SBC);(SCD))}=\widehat{IHK}$

Ta có: $\triangle SHB=\triangle SHM\ (c.g.c)$

$\Rightarrow HI = HK$

Áp dụng hệ thức lượng trong $\triangle SHM$ vuông tại $H$ đường cao $HK$ ta được:

$\quad \dfrac{1}{HK^2}=\dfrac{1}{SA^2} +\dfrac{1}{HM^2}$

$\Rightarrow HK =\dfrac{SA.HM}{\sqrt{SA^2 + HK^2}}$

$\Rightarrow HK =\dfrac{a\sqrt2.a}{\sqrt{2a^2 + a^2}}=\dfrac{a\sqrt6}{3}$

$\Rightarrow HI = HK =\dfrac{a\sqrt6}{3}$

Mặt khác:

Do $\triangle SHB=\triangle SHM\ (c.g.c)$

nên $\dfrac{SI}{SB}=\dfrac{SK}{SM}$

$\Rightarrow IK//BM$ (theo định lý $Thales$)

Ta được:

$\quad \dfrac{IK}{BM}=\dfrac{SK}{SM}$

$\Rightarrow IK = BM\cdot \dfrac{SK}{SM}$

$\Rightarrow IK =\sqrt{BC^2 + CM^2}\cdot \dfrac{SH^2}{SM^2}$

$\Rightarrow IK =\sqrt{BC^2 + CM^2}\cdot \dfrac{SH^2}{SH^2 + HM^2}$

$\Rightarrow IK = \sqrt{a^2 + a^2}\cdot \dfrac{2a^2}{2a^2 + a^2} = \dfrac{2a\sqrt2}{3}$

Áp dụng định lý $\cos$ vào $\triangle HIK$ ta được:

$\quad IK^2 = HI^2 + HK^2 - 2HI.HK.\cos\widehat{IHK}$

$\Rightarrow \cos\widehat{IHK}=\dfrac{HI^2 + HK^2 - IK^2}{2HI.HK}$

$\Rightarrow \cos\widehat{IHK}=\dfrac{2\cdot \dfrac{2a^2}{3} - \dfrac{8a^2}{9}}{2\cdot \dfrac{2a^2}{3}}$

$\Rightarrow \cos\widehat{IHK}=\dfrac13$

$\Rightarrow \widehat{IHK}=\arccos\dfrac13$

Vậy $\widehat{((SBC);(SCD))}=\arccos\dfrac13$