a.

\(SA\perp\left(ABCD\right)\Rightarrow AB\) là hình chiếu vuông góc của SB lên (ABCD)

\(\Rightarrow\widehat{SBA}=\left(SB;\left(ABCD\right)\right)\)

\(tan\widehat{SBA}=\dfrac{SA}{AB}=\dfrac{\sqrt{2}}{2}\Rightarrow\widehat{SBA}\approx35^016'\)

Tương tự \(SA\perp\left(ABCD\right)\Rightarrow\widehat{SCA}=\left(SC;\left(ABCD\right)\right)\)

\(AC=\sqrt{AD^2+DC^2}=a\sqrt{2}\)

\(\Rightarrow tan\widehat{SCA}=\dfrac{SA}{AC}=1\Rightarrow\widehat{SCA}=45^0\)

b.

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp AB\\AB\perp AD\end{matrix}\right.\) \(\Rightarrow AB\perp\left(SAD\right)\)

\(\Rightarrow\left(AH;\left(SAD\right)\right)=90^0-\left(AH;AB\right)=90^0-\widehat{HAB}\)

Gọi E là trung điểm AB \(\Rightarrow ADCE\) là hình vuông \(\Rightarrow\widehat{ACE}=45^0\)

Tam giác BCE vuông cân tại E (do \(EB=EC=a\)) nên \(\widehat{ECB}=45^0\)

\(\Rightarrow\widehat{ACB}=90^0\) hay \(BC\perp AC\Rightarrow BC\perp\left(SAC\right)\) (do \(SA\perp BC\))

\(\Rightarrow BC\perp AH\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp BH\)

Hay tam giác ABH vuông tại H 

\(AH=\dfrac{SA.AC}{\sqrt{SA^2+AC^2}}=a\)

\(\Rightarrow cos\widehat{HAB}=\dfrac{AH}{AB}=\dfrac{1}{2}\Rightarrow\widehat{HAB}=60^0\)

\(\Rightarrow\widehat{HAB}=60^0\Rightarrow\left(AH;\left(SAD\right)\right)=30^0\)

Theo cmt \(BC\perp\left(SAC\right)\Rightarrow\left(SB;\left(SAC\right)\right)=\widehat{BSC}\)

\(SC=\sqrt{SA^2+AC^2}=2a\) ; \(SB=\sqrt{SA^2+AB^2}=a\sqrt{6}\)

\(\Rightarrow cos\widehat{BSC}=\dfrac{SC}{SB}=\dfrac{\sqrt{6}}{3}\Rightarrow\widehat{BSC}\approx35^016'\)

a: \(AC=\sqrt{a^2+a^2}=a\sqrt{2}\)

(SC;(ABCD))=(CS;CA)=góc SCA

tan SCA=SA/AC=1/căn 2

=>góc SCA=35 độ

b:

Kẻ BH vuông góc AC tại H

(SB;SAC)=(SB;SH)=góc BSH

\(HB=\dfrac{a\cdot a}{a\sqrt{2}}=a\cdot\dfrac{\sqrt{2}}{2}\)

AH=AC/2=a*căn 2/2

=>\(SH=\sqrt{a^2+\dfrac{1}{2}a^2}=a\sqrt{\dfrac{3}{2}}\)

\(SH=\dfrac{a\sqrt{6}}{2};HB=\dfrac{a\sqrt{2}}{2};SB=a\sqrt{2}\)

\(cosBSH=\dfrac{SB^2+SH^2-BH^2}{2\cdot SB\cdot SH}=\dfrac{\sqrt{3}}{2}\)

=>góc BSH=30 độ

c: (SD;(SAB))=(SD;SA)=góc ASD

tan ASD=AD/AS=2

nên góc ASD=63 độ

Ta có: \(S_{ABCD}=\dfrac{\left(BC+AD\right).AB}{2}=\dfrac{3}{2}a^2\)

a, \(h=SA=AB.tan60^o=a\sqrt{3}\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.a\sqrt{3}=\dfrac{\sqrt{3}}{2}a^3\)

b, \(h=SA=AD.tan45^o=2a\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.2a=a^3\)

c, Dễ chứng minh được SC vuông góc với CD tại C \(\Rightarrow\widehat{SCA}=30^o\)

\(\Rightarrow h=SA=AC.tan30^o=AD.sin45^o.tan30^o=\dfrac{\sqrt{6}}{3}a\)

\(\Rightarrow V=\dfrac{1}{3}.S_{ABCD}.h=\dfrac{1}{3}.\dfrac{3}{2}a^2.\dfrac{\sqrt{6}}{3}a=\dfrac{\sqrt{6}}{6}a^3\)

Đáp án B

Dễ thấy 

Gọi H là trung điểm của AB 

Tam giác MHN vuông tại H, có 

Tam giác MHC vuông tại H, có 

Tam giác MNC, có  c o s M N C ^

Vậy cos(MN;(SAC)) =  sin M N C   ^ = 1 - cos 2 M N C ^ = 55 10

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\AB\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp AC\\BD\perp AC\left(\text{hai đường chéo hình vuông}\right)\end{matrix}\right.\)  \(\Rightarrow BD\perp\left(SAC\right)\)

\(BC\perp\left(SAB\right)\Rightarrow BC\perp AH\) ; mà \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\)

\(\left\{{}\begin{matrix}CD\perp\left(SAD\right)\Rightarrow CD\perp AK\\AK\perp SD\end{matrix}\right.\) \(\Rightarrow AK\perp\left(SCD\right)\)

\(\left\{{}\begin{matrix}AH\perp\left(SBC\right)\Rightarrow AH\perp SC\\AK\perp\left(SCD\right)\Rightarrow AK\perp SC\end{matrix}\right.\) \(\Rightarrow SC\perp\left(AHK\right)\Rightarrow SC\perp HK\)

Mặt khác theo tính đối xứng hình vuông \(\Rightarrow HK||BD\Rightarrow HK\perp AC\Rightarrow HK\perp\left(SAC\right)\)

\(AI\in\left(SAC\right)\Rightarrow HK\perp AI\)