Giải thích các bước giải:

1.Ta có $SA\perp (ABCD)\to \widehat{SB, ABCD}=\widehat{SBA}$

   Ta có $\tan\widehat{SBA}=\dfrac{SA}{AB}= \sqrt{3}\to \widehat{SBA}=60^o$

$\to \widehat{SB, ABCD}=60^o$

2.Tương tự câu a $\to \widehat{SD, ABCD}=60^o$

3.Ta có $ABCD$ là hình vuông cạnh $a\to AC=DB=a\sqrt{2}$

Mà $SA\perp ABCD\to \widehat{SC, ABCD}=\widehat{SCA}$

Ta có:

$\tan\widehat{SCA}=\dfrac{SA}{AC}=\sqrt{\dfrac32}$

$\to \widehat{SCA}=\arctan(\sqrt{\dfrac32})$

$\to \widehat{SC, ABCD}=\arctan(\sqrt{\dfrac32})$

4.Ta có $\tan\widehat{SOA}=\dfrac{SA}{AO}=\sqrt{6}$

$\to \widehat{SOA}=\arctan(\sqrt{6})$

$\to \widehat{SO, ABCD}=\arctan(\sqrt{6})$

5.Ta có $SB^2=SA^2+AB^2=4a^2\to SB=2a$

Tương tự $SC=2a$

Ta có $SC^2=SA^2+AC^2=a\sqrt{5}$

Ta có $AB\perp AD, SA\perp ABCD\to SA\perp AD$

$\to AB\perp SAD$

$\to \widehat{SB, SAD}=\widehat{BSA}=90^o-\widehat{SBA}=30^o$

6.Ta có $SA\perp ABCD\to SA\perp CD, CD\perp AD$

$\to CD\perp SAD\to CD\perp SD$

$\to\widehat{SC, SAD}= \widehat{CSD}$

Ta có $\tan\widehat{CSD}=\dfrac{CD}{SD}=\dfrac12\to \widehat{CSD}=\arctan\dfrac12$

$\to \widehat{SC, SAD}=\arctan\dfrac12$

7.Tương tự câu 5 $\widehat{SD, SAB}=30^o$

8.Tương tự câu 6$\to \widehat{SC, SAB}=\widehat{CSB}=\arctan\dfrac12$

9.Ta có $SA\perp ABCD\to SA\perp BD$

Mà $ABCD$ là hình vuông $\to AC\perp DB$

$\to BD\perp (SAC)$

$\to \widehat{SD, SAC}=\widehat{DSO}$

Ta có $\tan\widehat{DSO}=\dfrac{SO}{DO}=\sqrt{7}$

$\to \widehat{DSO}=\arctan\sqrt{7}$

$\to \widehat{SD, SAC}=\arctan\sqrt{7}$

10.Tương tự câu 9$\to \widehat{SB, SAC}=\arctan\sqrt{7}$