Cho hình chóp $S.ABCD$ có đáy là hình thang vuông tại A và...

Phương pháp: Gọi \)">a'\left( P \right).~\left( P \right)aa'.~ Gọi \)">E,FSC,AB.MENF~\left( ME//NF,\angle M=\angle F={{90}^{0}} \right)KNFAC,I~EKMNIMN\left( SAC \right)~\left\{ \begin{array}{*{35}{l}}
NC\bot AC \\
NC\bot SA \\
\end{array}\Rightarrow NC+(SAC) \right. C N(\text{SAC})\Rightarrow \angle \left( MN;(SAC) \right)=\angle NICNICC\Rightarrow \sin NIC=\dfrac{NC}{IN}NC=\dfrac{1}{2}CD=\dfrac{1}{2}.\sqrt{{{a}^{2}}+{{a}^{2}}}=\dfrac{a\sqrt{2}}{2},MN=\sqrt{M{{F}^{2}}+N{{F}^{2}}}=\sqrt{{{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{3a}{2} \right)}^{2}}}=\dfrac{a\sqrt{10}}{2},\dfrac{IN}{IM}=\dfrac{KN}{ME}=2\Rightarrow IN=\dfrac{2}{3}MN=\dfrac{2}{3}\cdot \dfrac{a\sqrt{10}}{2}=\dfrac{a\sqrt{10}}{3}\Rightarrow \sin NIC=\dfrac{NC}{IN}=\dfrac{3\sqrt{5}}{10}MN\dfrac{3\sqrt{5}}{10}.$