在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=根号2/2在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=√2/2 ①,求∠A.②,若边a=√5 ,S△=3 ,求边 b 和 c .

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在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=根号2/2在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=√2/2 ①,求∠A.②,若边a=√5 ,S△=3 ,求边 b 和 c .
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在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=根号2/2在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=√2/2 ①,求∠A.②,若边a=√5 ,S△=3 ,求边 b 和 c .
在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=根号2/2
在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),
且m×n=√2/2
①,求∠A.
②,若边a=√5 ,S△=3 ,求边 b 和 c .

在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=根号2/2在△ABC中,向量m=（cos A/2,-sinA/2),向量n=（cosA/2,sinA/2),且m×n=√2/2 ①,求∠A.②,若边a=√5 ,S△=3 ,求边 b 和 c .
(1)由倍角公式
mn
=(cos A/2,-sin A/2)(cos A/2,sin A/2)
=(cos A/2)^2-(sin A/2)^2
=cosA
=√2/2
所以 A=45度.
(2)由三角形面积公式：S=1/2bcsinA=3,而 sinA=√2/2,所以 bc=6√2.
由余弦定理：a^2=b^2+c^2-2bccosA,而 a=√5,bc=6√2,所以
b^2+c^2=17.结合 bc=6√2 可以解出 b = 3,c = 2√2 或者 b = 2√2,c=3.

(1）由已知m×n即向量m与n的向量积（即叉乘）的公式可得
m×n=√2/2=cos A/2*sinA/2-（-sinA/2*cosA/2）=2sinA/2*cosA/2=sinA
即sinA=√2/2
所以 A=π/4或3π/4
（2）由三角形面积公式：S=1/2bcsinA=3，而 sinA=√2/2
所以 bc=6√2.
由余弦定理：a...

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(1）由已知m×n即向量m与n的向量积（即叉乘）的公式可得
m×n=√2/2=cos A/2*sinA/2-（-sinA/2*cosA/2）=2sinA/2*cosA/2=sinA
即sinA=√2/2
所以 A=π/4或3π/4
（2）由三角形面积公式：S=1/2bcsinA=3，而 sinA=√2/2
所以 bc=6√2.
由余弦定理：a^2=b^2+c^2-2bccosA，而 a=√5，bc=6√2
所以当A=π/4时：b^2+c^2=17. 结合 bc=6√2
可以解出 b = 3，c = 2√2 或者 b = 2√2，c=3.
当A=3π/4时，b^2+c^2=-7（不合题意舍去）

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问题一只有一解 答案是45度 用二倍角公式就能算出

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