Đáp án đúng: C
Giải chi tiết:
Trong \(\left( {BCED} \right)\) kéo dài \(BC,\,\,DE\) cắt nhau tại \(F\).
Đăt \(CF = x\,\,\left( {x > 0} \right) \Rightarrow BF = a + x\).
Áp dụng định lí Cosin trong tam giác \(ABF\) ta có:
\(\begin{array}{l}\,\,\,\,\,\,A{F^2} = {a^2} + {\left( {a + x} \right)^2} - 2a.\left( {a + x} \right).\dfrac{1}{2}\\ \Leftrightarrow A{F^2} = 2{a^2} + 2ax + {x^2} - {a^2} - ax\\ \Leftrightarrow A{F^2} = {a^2} + ax + {x^2}\\ \Leftrightarrow AF = \sqrt {{a^2} + ax + {x^2}} \end{array}\)
Trong \(\left( {ADF} \right)\) kẻ \(DH \bot AF\,\,\left( {H \in AF} \right)\) ta có: \(\left\{ \begin{array}{l}AF \bot DH\\AF \bot BD\,\,\left( {gt} \right)\end{array} \right. \Rightarrow AF \bot \left( {BDH} \right) \Rightarrow AF \bot BH\).
\(\left\{ \begin{array}{l}\left( P \right) \equiv \left( {ADF} \right) \cap \left( {ABC} \right) \equiv \left( {ABF} \right) = AF\\DH \subset \left( {ADF} \right),\,\,DH \bot AF\\BH \subset \left( {ABF} \right),\,\,BH \bot AF\end{array} \right.\)\( \Rightarrow \angle \left( {\left( {ADF} \right);\left( {ABF} \right)} \right) = \angle \left( {DH;BH} \right) = \angle BHD = {60^0}\).
Ta lại có:
\(\begin{array}{l}{S_{ABF}} = \dfrac{1}{2}BH.AF = \dfrac{1}{2}AB.BF.\sin {60^0}\\ \Rightarrow BH = \dfrac{{a.\left( {a + x} \right).\dfrac{{\sqrt 3 }}{2}}}{{\sqrt {{a^2} + ax + {x^2}} }}\end{array}\)
Xét tam giác vuông \(BDH\) có: \(DH = \dfrac{{BH}}{{\cos {{60}^0}}} = \dfrac{{\sqrt 3 a.\left( {a + x} \right)}}{{\sqrt {{a^2} + ax + {x^2}} }}\)
\(\begin{array}{l} \Rightarrow {S_{ADF}} = \dfrac{1}{2}DH.AF\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}.\dfrac{{\sqrt 3 a.\left( {a + x} \right)}}{{\sqrt {{a^2} + ax + {x^2}} }}.\sqrt {{a^2} + ax + {x^2}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}.\sqrt 3 a.\left( {a + x} \right)\end{array}\)
Ta có:
\(\begin{array}{l}\dfrac{{{S_{ADE}}}}{{{S_{ADF}}}} = \dfrac{{DE}}{{DF}} = \dfrac{{BC}}{{BF}} = \dfrac{a}{{a + x}}\\ \Rightarrow {S_{ADE}} = \dfrac{a}{{a + x}}.\dfrac{1}{2}\sqrt 3 a\left( {a + x} \right) = \dfrac{{\sqrt 3 {a^2}}}{2}\\ \Rightarrow \dfrac{1}{2}.AD.AE.\sin \beta = \dfrac{{\sqrt 3 {a^2}}}{2}\\ \Rightarrow \dfrac{{a\sqrt 6 }}{2}.a\sqrt 3 .\sin \beta = \sqrt 3 {a^2}\\ \Leftrightarrow \sin \beta = \dfrac{{\sqrt 6 }}{3} = \dfrac{6}{{3\sqrt 6 }} = \dfrac{2}{{\sqrt 6 }}\end{array}\)
Chọn C.
