Đáp án:
$\sin\widehat{MAN}=\dfrac{\sqrt2}{2}$
Giải thích các bước giải:
Ta có:
$DM = \dfrac13CD \to DM =\dfrac a3;\, CM =\dfrac{2a}{3}$
$BN = NC =\dfrac12BC =\dfrac a2$
Áp dụng định lý Pytago ta được:
$+)\quad AM^2 = AD^2 + DM^2 = a^2 + \dfrac{a^2}{9}$
$\to AM = \dfrac{a\sqrt{10}}{3}$
$+)\quad AN^2 = AB^2 + BN^2=a^2 + \dfrac{a^2}{4}$
$\to AN = \dfrac{a\sqrt5}{2}$
Mặt khác:
$S_{AMN}=S_{ABCD}-S_{ABN} -S_{CMN} - S_{ADM}$
$\to S_{AMN}=AB^2 - \dfrac12AB.BM -\dfrac12CN.CM -\dfrac12AD.DM$
$\to S_{AMN}=a^2 -\dfrac12\cdot\dfrac a2\cdot a -\dfrac12\cdot\dfrac a2\cdot\dfrac{2a}{3} -\dfrac12\dfrac a3\cdot a$
$\to S_{AMN}=\dfrac{5a^2}{12}$
Ta có:
$S_{AMN}=\dfrac12AM.AN.\sin\widehat{MAN}$
$\to \sin\widehat{MAN}=\dfrac{2S_{AMN}}{AM.AN}$
$\to \sin\widehat{MAN}=\dfrac{2\cdot\dfrac{5a^2}{12}}{\dfrac{a\sqrt{10}}{3}\cdot\dfrac{a\sqrt5}{2}}$
$\to \sin\widehat{MAN}=\dfrac{\sqrt2}{2}$
