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2022年高考数学新高考Ⅱ-18

  2022-12-16 17:32:55  

(12分)记$\Delta ABC$的内角$A$,$B$,$C$的对边分别为$a$,$b$,$c$,分别以$a$,$b$,$c$为边长的三个正三角形的面积依次为$S_{1}$,$S_{2}$,$S_{3}$.已知$S_{1}-S_{2}+S_{3}=\dfrac{\sqrt{3}}{2}$,$\sin B=\dfrac{1}{3}$.
(1)求$\Delta ABC$的面积;
(2)若$\sin A\sin C=\dfrac{\sqrt{2}}{3}$,求$b$.
分析:(1)根据$S_{1}-S_{2}+S_{3}=\dfrac{\sqrt{3}}{2}$,求得$a^{2}-b^{2}+c^{2}=2$,由余弦定理求得$ac$的值,根据$S=\dfrac{1}{2}ac\sin B$,求$\Delta ABC$面积.
(2)由正弦定理得$\therefore a=\dfrac{b\sin A}{\sin B}$,$c=\dfrac{b\sin C}{\sin B}$,且$ac=\dfrac{3\sqrt{2}}{4}$,求解即可.
解:(1)$S_{1}=\dfrac{1}{2}a^{2}\sin 60^\circ =\dfrac{\sqrt{3}}{4}a^{2}$,
$S_{2}=\dfrac{1}{2}b^{2}\sin 60^\circ =\dfrac{\sqrt{3}}{4}b^{2}$,
$S_{3}=\dfrac{1}{2}c^{2}\sin 60^\circ =\dfrac{\sqrt{3}}{4}c^{2}$,
$\because S_{1}-S_{2}+S_{3}=\dfrac{\sqrt{3}}{4}a^{2}-\dfrac{\sqrt{3}}{4}b^{2}+\dfrac{\sqrt{3}}{4}c^{2}=\dfrac{\sqrt{3}}{2}$,
解得:$a^{2}-b^{2}+c^{2}=2$,
$\because \sin B=\dfrac{1}{3}$,$a^{2}-b^{2}+c^{2}=2 > 0$,即$\cos B > 0$,
$\therefore \cos B=\dfrac{2\sqrt{2}}{3}$,
$\therefore \cos B=\dfrac{a^{2}+c^{2}-b^{2}}{2ac}=\dfrac{2\sqrt{2}}{3}$,
解得:$ac=\dfrac{3\sqrt{2}}{4}$,
$S_{\Delta ABC}=\dfrac{1}{2}ac\sin B=\dfrac{\sqrt{2}}{8}$.
$\therefore \Delta ABC$的面积为$\dfrac{\sqrt{2}}{8}$.
(2)由正弦定理得:$\dfrac{b}{\sin B}=\dfrac{a}{\sin A}=\dfrac{c}{\sin C}$,
$\therefore a=\dfrac{b\sin A}{\sin B}$,$c=\dfrac{b\sin C}{\sin B}$,
由(1)得$ac=\dfrac{3\sqrt{2}}{4}$,
$\therefore ac=\dfrac{b\sin A}{\sin B}\cdot \dfrac{b\sin C}{\sin B}=\dfrac{3\sqrt{2}}{4}$
已知,$\sin B=\dfrac{1}{3}$,$\sin A\sin C=\dfrac{\sqrt{2}}{3}$,
解得:$b=\dfrac{1}{2}$.
点评:本题考查利用正余弦定理解三角形,需灵活运用正余弦定理公式.

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