Question

在三角形ABC中a、b、c分别为角A、B、C的对边，m=﹙2a－c,b﹚,n=﹙cosC,cosB﹚,且m∥n，﹙1﹚，求角B的大小

在三角形ABC中a、b、c分别为角A、B、C的对边，m=﹙2a－c,b﹚,n=﹙cosC,cosB﹚,且m∥n，﹙1﹚，求角B的大小；﹙2﹚若b=根号3，求a＋b的最大值。

Answer 1

因为m//n且m=(2a-c，b)，n=(cosC，cosB)
则( 2a-c )cosB-bcosC=0
正弦定梁知理a/sinA=b/sinB=c/sinC,则
(2sinA-sinC)cosB-sinBcosC=0
2sinAcosB-sin(B+C) =0
2sinAcosB-sinA=0
所虚誉以cosB=1/2
即B=60°

2）又因 b=√3
则用余弦定理得：3=a²+c²-2accos60°
即a²+c²-ac=3
则(a+c)²=a²+c²-ac+3ac=3+3ac<=3+3(a²+c²)/2
当a=c时，(a+c)²取得最大值
则 a²=3 即a=c=√3
所以橡誉消a+c=2√3

Answer 2

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Answer 3

用M=λN 求出cosB的值 就是所求角B的大小了！