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Câu hỏi: Cho hình chóp $S.ABCD$ có đáy $ABCD$ là hình chữ nhật, $SA=a\sqrt{5},AB=4a,AD=a\sqrt{3}$. Điểm $H$ nằm trên cạnh $AB$ thỏa mãn $AH=\dfrac{1}{3}HB$, hai mặt phẳng $\left( SHC \right)$ và $\left( SHD \right)$ cùng vuông góc với mặt phẳng đáy. Cosin góc giữa $SD$ và $\left( SBC \right)$ bằng
A. $\sqrt{\dfrac{5}{12}}$.
B. $\sqrt{\dfrac{5}{13}}$.
C. $\sqrt{\dfrac{4}{13}}$.
D. $\dfrac{\sqrt{3}}{3}$.
Ta có: $\sin \widehat{\left( SD;\left( SBC \right) \right)}=\dfrac{d\left( D;\left( SBC \right) \right)}{SD}$
Do $AD//BC\Rightarrow AD//\left( SBC \right)$
$\Rightarrow d\left( D;\left( SBC \right) \right)=d\left( A;\left( SBC \right) \right)$
Lại có: $AB=\dfrac{4}{3}HB\Rightarrow d\left( A;\left( SBC \right) \right)=\dfrac{4}{3}d\left( H;\left( SBC \right) \right)$
Do $\left\{ \begin{aligned}
& HB\bot BC \\
& SH\bot BC \\
\end{aligned} \right.\Rightarrow BC\bot \left( SBH \right)$
Dựng $HE\bot SB\Rightarrow HE\bot \left( SBC \right)$
Ta có: $HA=a,HB=3a\Rightarrow SH=\sqrt{S{{A}^{2}}-H{{A}^{2}}}=2a,d\left( H;\left( SBC \right) \right)=HE=\dfrac{HB.SH}{\sqrt{H{{B}^{2}}+S{{H}^{2}}}}=\dfrac{6a}{\sqrt{13}}$.
$SD=\sqrt{S{{H}^{2}}+H{{D}^{2}}}=\sqrt{S{{H}^{2}}+H{{A}^{2}}+A{{D}^{2}}}=2a\sqrt{2}$.
Suy ra $\sin \widehat{\left( SD;\left( SBC \right) \right)}=\dfrac{\dfrac{4}{3}HE}{SD}=\dfrac{2\sqrt{26}}{13}\Rightarrow \cos \widehat{\left( SD;\left( SBC \right) \right)}=\sqrt{1-{{\sin }^{2}}\widehat{\left( SD;\left( SBC \right) \right)}}=\sqrt{\dfrac{5}{13}}$.
A. $\sqrt{\dfrac{5}{12}}$.
B. $\sqrt{\dfrac{5}{13}}$.
C. $\sqrt{\dfrac{4}{13}}$.
D. $\dfrac{\sqrt{3}}{3}$.
Ta có: $\sin \widehat{\left( SD;\left( SBC \right) \right)}=\dfrac{d\left( D;\left( SBC \right) \right)}{SD}$
Do $AD//BC\Rightarrow AD//\left( SBC \right)$
$\Rightarrow d\left( D;\left( SBC \right) \right)=d\left( A;\left( SBC \right) \right)$
Lại có: $AB=\dfrac{4}{3}HB\Rightarrow d\left( A;\left( SBC \right) \right)=\dfrac{4}{3}d\left( H;\left( SBC \right) \right)$
Do $\left\{ \begin{aligned}
& HB\bot BC \\
& SH\bot BC \\
\end{aligned} \right.\Rightarrow BC\bot \left( SBH \right)$
Dựng $HE\bot SB\Rightarrow HE\bot \left( SBC \right)$
Ta có: $HA=a,HB=3a\Rightarrow SH=\sqrt{S{{A}^{2}}-H{{A}^{2}}}=2a,d\left( H;\left( SBC \right) \right)=HE=\dfrac{HB.SH}{\sqrt{H{{B}^{2}}+S{{H}^{2}}}}=\dfrac{6a}{\sqrt{13}}$.
$SD=\sqrt{S{{H}^{2}}+H{{D}^{2}}}=\sqrt{S{{H}^{2}}+H{{A}^{2}}+A{{D}^{2}}}=2a\sqrt{2}$.
Suy ra $\sin \widehat{\left( SD;\left( SBC \right) \right)}=\dfrac{\dfrac{4}{3}HE}{SD}=\dfrac{2\sqrt{26}}{13}\Rightarrow \cos \widehat{\left( SD;\left( SBC \right) \right)}=\sqrt{1-{{\sin }^{2}}\widehat{\left( SD;\left( SBC \right) \right)}}=\sqrt{\dfrac{5}{13}}$.
Đáp án B.