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nga132

T

trantien.hocmai

$\text{giải} \\
\widehat{[SB,(ABCD)]}=\widehat{SBA}=60^o \\$
$$SA=AB.\tan \widehat{SBA}=a.\tan 60^o=a\sqrt{3} \\
SM=SA-AM=a\sqrt{3}-\dfrac{a\sqrt{3}}{3}=\dfrac{2a\sqrt{3}}{3} \\
\rightarrow \dfrac{SM}{SA}=\dfrac{SN}{SD}=\dfrac{2}{3}$$
 
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