Đáp án: $[B, SA, C]=\arccos(\dfrac{2617}{102\sqrt{195}})$

Giải thích các bước giải:

Kẻ $CE\perp, SA, BF\perp SA$

Qua $F$ kẻ $FG//CE$

$\to [B, SA, C]=\widehat{CE, BF}=\widehat{GF, FB}=\widehat{BFG}$$

VÌ $SC\perp (ABCD)$

$\to SC\perp AC, SC\perp BC$

Ta có: $AB=2a\sqrt2$

$\to SA=\sqrt{SC^2+CA^2}=\sqrt{(3a)^2+(2\sqrt2a)^2}=a\sqrt{17}$

      $SB=\sqrt{SC^2+BC^2}=\sqrt{(3a)^2+(2a)^2}=a\sqrt{13}$

Vì $SC\perp AB, AB\perp CB$

$\to AB\perp (SBC)$

$\to AB\perp SB$

$\to \dfrac1{BF^2}=\dfrac1{SB^2}+\dfrac1{SA^2}$

$\to \dfrac1{BF^2}=\dfrac1{(a\sqrt{13})^2}+\dfrac1{(a\sqrt{17})^2}$

$\to BF=\dfrac{\sqrt{221}a}{\sqrt{30}}$

$\to AF=\dfrac{AB^2}{SA}=\dfrac{4a^2}{a\sqrt{17}}=\dfrac{2a}{\sqrt{17}}$

Ta có: $\Delta SCA$ vuông tại $C, CE\perp SA$

$\to AE=\dfrac{CA^2}{AS}=\dfrac{(2\sqrt2a)^2}{a\sqrt{17}}=\dfrac{8a}{\sqrt{17}}$

$\to CE=\sqrt{AC^2-AE^2}=\sqrt{(2a\sqrt2)^2-(\dfrac{8a}{\sqrt{17}})^2}=\dfrac{6\sqrt{2}a}{\sqrt{17}}$

$\to FG=\dfrac{AF}{AE}\cdot CE$

$\to FG=\dfrac{\dfrac{2a}{\sqrt{17}}}{\dfrac{8a}{\sqrt{17}}}\cdot \dfrac{6\sqrt{2}a}{\sqrt{17}}$

$\to FG=\dfrac{3a}{\sqrt{34}}$

Ta có: 

$AG=\dfrac{AF}{AE}\cdot AC=\dfrac{\dfrac{2a}{\sqrt{17}}}{\dfrac{8a}{\sqrt{17}}}\cdot2a\sqrt2=\dfrac{a}{\sqrt{2}}$

Mà $ABCD$ là hình vuông

Gọi $AC\perp BD=O\to O$ là trung điểm $AC, DB$

$\to OG=OA-AG=a\sqrt2-\dfrac{a}{\sqrt{2}}=\dfrac{a}{\sqrt{2}}$

$\to BG=\sqrt{BO^2+OG^2}=\sqrt{(a\sqrt2)^2+(\dfrac{a}{\sqrt{2}})^2}=\dfrac{\sqrt{5}a}{\sqrt{2}}$

$\to \cos\widehat{BFG}=\dfrac{FB^2+FG^2-BG^2}{2FB\cdot FG}$

$\to \cos\widehat{BFG}=\dfrac{(\dfrac{\sqrt{221}a}{\sqrt{30}})^2+(\dfrac{3a}{\sqrt{34}})^2-(\dfrac{\sqrt{5}a}{\sqrt{2}})^2}{2\cdot \dfrac{\sqrt{221}a}{\sqrt{30}}\cdot \dfrac{3a}{\sqrt{34}}}$

$\to\cos\widehat{BFG}=\dfrac{2617}{102\sqrt{195}}$

$\to\widehat{BFG}=\arccos(\dfrac{2617}{102\sqrt{195}})$

$\to [B, SA, C]=\arccos(\dfrac{2617}{102\sqrt{195}})$