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2022年高考数学北京16

  2022-12-16 20:10:06  

(13分)在$\Delta ABC$中,$\sin 2C=\sqrt{3}\sin C$.
(Ⅰ)求$\angle C$;
(Ⅱ)若$b=6$,且$\Delta ABC$的面积为$6\sqrt{3}$,求$\Delta ABC$的周长.
分析:(Ⅰ)根据二倍角公式化简可得$\cos C$,进一步计算可得角$C$;(Ⅱ)根据三角形面积求得$a$,再根据余弦定理求得$c$,相加可得三角形的周长.
解答:解:(Ⅰ)$\because \sin 2C=\sqrt{3}\sin C$,
$\therefore 2\sin C\cos C=\sqrt{3}\sin C$,
又$\sin C\ne 0$,$\therefore 2\cos C=\sqrt{3}$,
$\therefore \cos C=\dfrac{\sqrt{3}}{2}$,$\because 0 < C < \pi$,
$\therefore C=\dfrac{\pi }{6}$;
(Ⅱ)$\because \Delta ABC$的面积为$6\sqrt{3}$,
$\therefore$$\dfrac{1}{2}ab\sin C=6\sqrt{3}$,
又$b=6$,$C=\dfrac{\pi }{6}$,
$\therefore$$\dfrac{1}{2}\times a\times 6\times \dfrac{1}{2}=6\sqrt{3}$,
$\therefore a=4\sqrt{3}$,
又$\cos C=\dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$,
$\therefore$$\dfrac{\sqrt{3}}{2}=\dfrac{(4\sqrt{3})^{2}+{6}^{2}-{c}^{2}}{2\times 4\sqrt{3}\times 6}$,
$\therefore c=2\sqrt{3}$,
$\therefore a+b+c=6+6\sqrt{3}$,
$\therefore \Delta ABC$的周长为$6+6\sqrt{3}$.
点评:本题考查了三角形面积公式和余弦定理的应用,属于中档题.

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