Giải thích các bước giải:

Kẻ $BH\perp AC$

Vì $SB\perp (ABCD)\to SB\perp AC$

$\to AC\perp (SBH)$

$\to AC\perp SH$

$\to \widehat{SAC, ABCD}=\widehat{SHB}$

Ta có: $\Delta ABC$ vuông tại $B, BH\perp AC$

$\to \dfrac1{HB^2}=\dfrac1{AB^2}+\dfrac1{BC^2}$

$\to \dfrac1{HB^2}=\dfrac1{(3a)^2}+\dfrac1{(4a)^2}$

$\to HB=\dfrac{12}5a$

$\to \tan\alpha=\tan\widehat{SHB}=\dfrac{SB}{HB}=\dfrac{2a}{\dfrac{12}5a}=\dfrac56$