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2021年高考数学新高考Ⅰ-19

  2021-06-14 22:00:57  

(12分)记$\Delta ABC$的内角$A$,$B$,$C$的对边分别为$a$,$b$,$c$.已知$b^{2}=ac$,点$D$在边$AC$上,$BD\sin \angle ABC=a\sin C$.
(1)证明:$BD=b$;
(2)若$AD=2DC$,求$\cos \angle ABC$.
分析:(1)利用正弦定理求解;
(2)要能找到隐含条件:$\angle BDA$和$\angle BDC$互补,从而列出等式关系求解.
解:(1)证明:由正弦定理知,$\dfrac{b}{\sin \angle ABC}=\dfrac{c}{\sin \angle ACB}=2R$,
$\therefore b=2R\sin \angle ABC$,$c=2R\sin \angle ACB$,
$\because b^{2}=ac$,
$\therefore b\cdot 2R\sin \angle ABC=a\cdot 2R\sin \angle ACB$,
即$b\sin \angle ABC=a\sin C$,
$\because BD\sin \angle ABC=a\sin C$.
$\therefore BD=b$;
(2)由(1)知$BD=b$,
$\because AD=2DC$,
$\therefore AD=\dfrac{2}{3}b$,$DC=\dfrac{1}{3}b$,
在$\Delta ABD$中,由余弦定理知,$\cos \angle BDA=\dfrac{B{D}^{2}+A{D}^{2}-A{B}^{2}}{2BD\cdot AD}=\dfrac{{b}^{2}+(\dfrac{2}{3}{b)}^{2}-{c}^{2}}{2b\cdot \dfrac{2}{3}b}=\dfrac{13{b}^{2}-9{c}^{2}}{12{b}^{2}}$,
在$\Delta CBD$中,由余弦定理知,$\cos \angle BDC=\dfrac{B{D}^{2}+C{D}^{2}-B{C}^{2}}{2BD\cdot CD}=\dfrac{{b}^{2}+(\dfrac{1}{3}b)^{2}-{a}^{2}}{2b\cdot \dfrac{1}{3}b}=\dfrac{10{b}^{2}-9{a}^{2}}{6{b}^{2}}$,
$\because \angle BDA+\angle BDC=\pi$,
$\therefore \cos \angle BDA+\cos \angle BDC=0$,
即$\dfrac{13{b}^{2}-9{c}^{2}}{12{b}^{2}}+\dfrac{10{b}^{2}-9{a}^{2}}{6{b}^{2}}=0$,
得$11b^{2}=3c^{2}+6a^{2}$,
$\because b^{2}=ac$,
$\therefore 3c^{2}-11ac+6a^{2}=0$,
$\therefore c=3a$或$c=\dfrac{2}{3}a$,
在$\Delta ABC$中,由余弦定理知,$\cos \angle ABC=\dfrac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}=\dfrac{{a}^{2}+{c}^{2}-ac}{2ac}$,
当$c=3a$时,$\cos \angle ABC=\dfrac{7}{6}>1$(舍$)$;
当$c=\dfrac{2}{3}a$时,$\cos \angle ABC=\dfrac{7}{12}$;
综上所述,$\cos \angle ABC=\dfrac{7}{12}$.
点评:本题考查正弦定理及余弦定理的内容,是一道好题.

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