题目
21.(2024·全国)记 Delta ABC 内角A、B C的对边分别为a,b,c,已知 sin C=sqrt (2)cos B 。-|||-^2+(b)^2-(c)^2=sqrt (2)ab-|||-(1)求B;-|||-(2)若 Delta ABC 的面积为 +sqrt (3) ,求c。
题目解答
答案
【答案】(1)$\dfrac{\mathrm{\text{π}}}{3}$;(2)$2\sqrt{2}$.【解析】(1)由余弦定理得$cosC=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{\sqrt{2}}{2},C\in \left(0,\pi \right),\therefore C=\dfrac{\mathrm{\text{π}}}{4}$,又$\sin C=\sqrt{2}cosB,B\in \left(0,\mathrm{\text{π}}\right)\therefore B=\dfrac{\mathrm{\text{π}}}{3}$
(2)$\because 3+\sqrt{3}={S}_{\mathrm{\triangle }ABC}=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}bc\sin \left(\dfrac{2\pi }{3}-\dfrac{\pi }{4}\right)$$\therefore bc=4\sqrt{6}$又由正弦定理知：$\dfrac{b}{\sin B}=\dfrac{c}{\sin {\displaystyle C}}$,且$B=\dfrac{\mathrm{\text{π}}}{3}，C=\dfrac{\mathrm{\text{π}}}{4}$
$\therefore 4\sqrt{6}=bc=\dfrac{\sqrt{3}}{\sqrt{2}}{c}^{2},\therefore c=2\sqrt{2}$
解析
步骤 1：利用余弦定理求解C
由余弦定理，我们有：
\[ \cos C = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2ab} \]
代入已知条件 ${a}^{2} + {b}^{2} - {c}^{2} = \sqrt{2}ab$，得到：
\[ \cos C = \frac{\sqrt{2}ab}{2ab} = \frac{\sqrt{2}}{2} \]
因此，$C = \frac{\pi}{4}$，因为 $C \in (0, \pi)$。
步骤 2：利用已知条件求解B
已知 $\sin C = \sqrt{2} \cos B$，代入 $C = \frac{\pi}{4}$，得到：
\[ \sin \frac{\pi}{4} = \sqrt{2} \cos B \]
\[ \frac{\sqrt{2}}{2} = \sqrt{2} \cos B \]
\[ \cos B = \frac{1}{2} \]
因此，$B = \frac{\pi}{3}$，因为 $B \in (0, \pi)$。
步骤 3：利用三角形面积公式求解c
已知 $\Delta ABC$ 的面积为 $3 + \sqrt{3}$，利用三角形面积公式：
\[ S_{\Delta ABC} = \frac{1}{2}bc \sin A \]
代入已知条件，得到：
\[ 3 + \sqrt{3} = \frac{1}{2}bc \sin \left(\frac{2\pi}{3} - \frac{\pi}{4}\right) \]
\[ 3 + \sqrt{3} = \frac{1}{2}bc \sin \left(\frac{5\pi}{12}\right) \]
\[ 3 + \sqrt{3} = \frac{1}{2}bc \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \]
\[ 3 + \sqrt{3} = \frac{\sqrt{6} + \sqrt{2}}{8}bc \]
\[ bc = \frac{8(3 + \sqrt{3})}{\sqrt{6} + \sqrt{2}} \]
\[ bc = 4\sqrt{6} \]
由正弦定理，我们有：
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
代入 $B = \frac{\pi}{3}$，$C = \frac{\pi}{4}$，得到：
\[ \frac{b}{\sin \frac{\pi}{3}} = \frac{c}{\sin \frac{\pi}{4}} \]
\[ \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{\sqrt{2}}{2}} \]
\[ b = \frac{\sqrt{3}}{\sqrt{2}}c \]
代入 $bc = 4\sqrt{6}$，得到：
\[ \frac{\sqrt{3}}{\sqrt{2}}c^2 = 4\sqrt{6} \]
\[ c^2 = \frac{4\sqrt{6} \cdot \sqrt{2}}{\sqrt{3}} \]
\[ c^2 = 8 \]
\[ c = 2\sqrt{2} \]
由余弦定理，我们有：
\[ \cos C = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2ab} \]
代入已知条件 ${a}^{2} + {b}^{2} - {c}^{2} = \sqrt{2}ab$，得到：
\[ \cos C = \frac{\sqrt{2}ab}{2ab} = \frac{\sqrt{2}}{2} \]
因此，$C = \frac{\pi}{4}$，因为 $C \in (0, \pi)$。
步骤 2：利用已知条件求解B
已知 $\sin C = \sqrt{2} \cos B$，代入 $C = \frac{\pi}{4}$，得到：
\[ \sin \frac{\pi}{4} = \sqrt{2} \cos B \]
\[ \frac{\sqrt{2}}{2} = \sqrt{2} \cos B \]
\[ \cos B = \frac{1}{2} \]
因此，$B = \frac{\pi}{3}$，因为 $B \in (0, \pi)$。
步骤 3：利用三角形面积公式求解c
已知 $\Delta ABC$ 的面积为 $3 + \sqrt{3}$，利用三角形面积公式：
\[ S_{\Delta ABC} = \frac{1}{2}bc \sin A \]
代入已知条件，得到：
\[ 3 + \sqrt{3} = \frac{1}{2}bc \sin \left(\frac{2\pi}{3} - \frac{\pi}{4}\right) \]
\[ 3 + \sqrt{3} = \frac{1}{2}bc \sin \left(\frac{5\pi}{12}\right) \]
\[ 3 + \sqrt{3} = \frac{1}{2}bc \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \]
\[ 3 + \sqrt{3} = \frac{\sqrt{6} + \sqrt{2}}{8}bc \]
\[ bc = \frac{8(3 + \sqrt{3})}{\sqrt{6} + \sqrt{2}} \]
\[ bc = 4\sqrt{6} \]
由正弦定理，我们有：
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
代入 $B = \frac{\pi}{3}$，$C = \frac{\pi}{4}$，得到：
\[ \frac{b}{\sin \frac{\pi}{3}} = \frac{c}{\sin \frac{\pi}{4}} \]
\[ \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{\sqrt{2}}{2}} \]
\[ b = \frac{\sqrt{3}}{\sqrt{2}}c \]
代入 $bc = 4\sqrt{6}$，得到：
\[ \frac{\sqrt{3}}{\sqrt{2}}c^2 = 4\sqrt{6} \]
\[ c^2 = \frac{4\sqrt{6} \cdot \sqrt{2}}{\sqrt{3}} \]
\[ c^2 = 8 \]
\[ c = 2\sqrt{2} \]