a, \(\Delta ABC\) có \(\widehat{C}=90^o\).

Áp dụng pytago có: \(AB=\sqrt{AC^2+BC^2}=\sqrt{\left(12a\right)^2+\left(5a\right)^2}=13a\)

\(\Delta ABC\) có \(\widehat{C}=90^o\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\sin B=\dfrac{AC}{AB}=\dfrac{12a}{13a}=\dfrac{12}{13}\\cosB=\dfrac{BC}{AB}=\dfrac{5a}{13a}=\dfrac{5}{13}\end{matrix}\right.\)

Ta có: \(\dfrac{sinB+cosB}{sinB-cosB}=\dfrac{\dfrac{12}{13}+\dfrac{5}{13}}{\dfrac{12}{13}-\dfrac{5}{13}}=\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}=\dfrac{17}{7}\)

b, Có S_ABCD= \(\dfrac{CH.AB}{2}=\dfrac{CB.AC}{2}\Rightarrow CH.AB=BC.AC\Rightarrow CH=\dfrac{AC.BC}{AB}=\dfrac{12a.5a}{13a}=\dfrac{60a}{13}\approx4,615a\)

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

Mà CD là giao tuyến (SCD) và (ABCD)

\(\Rightarrow\widehat{SDA}\) là góc giữa (SCD) và (ABCD)

\(\Rightarrow\widehat{SDA}=60^0\Rightarrow SA=AD.tan60^0=3a\)

a.

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

Mà \(BC=\left(SBC\right)\cap\left(ABCD\right)\Rightarrow\widehat{SBA}\) là góc giữa (SBC) và (ABCD)

\(tan\widehat{SBA}=\dfrac{SA}{AB}=3\Rightarrow\widehat{SBA}=...\)

b.

Từ A kẻ \(AE\perp BD\)

\(SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\)

\(\Rightarrow BD\perp\left(SAE\right)\Rightarrow\widehat{SEA}\) là góc giữa (SBD) và (ABCD)

Hệ thức lượng: \(\dfrac{1}{AE^2}=\dfrac{1}{AB^2}+\dfrac{1}{AD^2}\Rightarrow AE=\dfrac{a\sqrt{3}}{2}\)

\(tan\widehat{SEA}=\dfrac{SA}{AE}=2\sqrt{3}\Rightarrow\widehat{SEA}=...\)

Chọn D

Theo định lý cosin ta có 

\(AD^2=AM^2+MD^2-2.MA.MD.cos\widehat{ÀMD}\)

Xé \(\Delta ABM\)có \(BM=\frac{a}{2}\)

 \(AM=\sqrt{AB^2+BM^2}=\sqrt{a^2+\left(\frac{a}{2}\right)^2}=\frac{\sqrt{5}a}{2}\)

Xét \(\Delta DCM\)có \(CM=\frac{a}{2}\)

\(\Rightarrow DM=\sqrt{DC^2+CM^2}=\sqrt{a^2+\left(\frac{a}{2}\right)^2}=\frac{\sqrt{5}a}{2}\)

\(\Rightarrow\cos\widehat{AMD}=\frac{AM^2+MD^2-AD^2}{2.MA.MD}=\frac{\frac{5a^2}{4}+\frac{5a^2}{4}-a^2}{\frac{\sqrt{5}a}{2}.\frac{\sqrt{5}a}{2}}=\frac{3}{5}\)

Vậy \(\cos\widehat{AMD}=\frac{3}{5}\)

cám ơn bạn nha

Chọn A.

Gắn tọa độ Oxyz, với A(0;0;0), B(1;0;0), D(0;3;0), S(0;0;1)

Khi đó C ( 1 ; 3 ; 0 ) ⇒  Trung điểm M của BC là M ( 1 ; 3 2 ; 0 ) .  

Ta có

SM → = ( 1 ; 3 2 ; - 1 ) , SD →   = ( 0 ; 3 ; - 1 ) ⇒ [ SM →   ; SD → ] = ( 3 2 ; 1 ; 3 ) .  

Suy ra n ⃗ ( SDM ) = ( 3 2 ; 1 ; 3 )  mà n ⃗ ( ABCD ) = n ⃗ ( Oxy ) = ( 0 ; 0 ; 1 ) ,  

ta được

cos ( SDM ^ ) ;   ( ABCD )   =   n → ( SDM ) . n → ( ABCD ) n → ( SDM ) . n → ( ABCD ) = 6 7 .