Giải thích các bước giải:

Bài 2:

a.Gọi $SE\perp CD,E\in CD$

Vì $SH\perp ABCD\to SH\perp CD\to CD\perp (SHE)\to CD\perp HE$

$\to \widehat{SCD,ABCD}=\widehat{SEH}$

Ta có : $BH=2AH,AB=a\to BH=\dfrac 23a, AH=\dfrac 13a$

$\to CH=\sqrt{BH^2+BC^2}=\dfrac{a\sqrt{13}}{3}$

Mà $\widehat{SC,ABCD}=45^o\to SH=CH=\dfrac{a\sqrt{13}}{3}$

Ta có : $AD\perp HD\to HD=\sqrt{AH^2+AD^2}=a.\dfrac{\sqrt{229}}{6},CD=a.\dfrac{\sqrt{13}}{2}$

$\to S_{HCD}=a^2$ tính theo công thức herong

$\to\dfrac 12 HE.CD=1\to HE=\dfrac{4a}{\sqrt{13}}$ 

$\to\tan\widehat{SC,ABCD}=\dfrac{SH}{HE}=\dfrac{\dfrac{a\sqrt{13}}{3}}{\dfrac{4a}{\sqrt{13}}}=\dfrac{13}{12}$

$\to \widehat{SC,ABCD}=\arctan\dfrac{13}{12}$

b.Gọi $IF\perp AB=F\to IF//SH\to\dfrac{IF}{SH}=\dfrac{AI}{AS}=\dfrac 13$

$\to IF=\dfrac{a\sqrt{13}}{9}$

Lại có : $\dfrac{AF}{AH}=\dfrac{AI}{AS}=\dfrac 13\to AF=\dfrac 13AH=\dfrac 13.\dfrac 13AB=\dfrac 19AB$

$\to BF=\dfrac 89 AB=\dfrac 89a$

Do $IF//SH, SH\perp (ABCD)\to IF\perp (ABCD)$

Mà $BC\perp AB,SH\perp BC\to BC\perp (SAB)\to \widehat{IBC,ABCD}=\widehat{IBF}$

$\to\tan\widehat{IBF}=\dfrac{IF}{BF}=\dfrac{\dfrac{a\sqrt{13}}{9}}{\dfrac{8}{9}a}=\dfrac{\sqrt{13}}{8}$

$\to \widehat{IBC,ABCD}=\arctan\dfrac{\sqrt{13}}{8}$