Question

在三角形ABC中，三个角A、B、C的对边边长分别为a=3，b=4，c=6，则bccosA+accosB+abcosC的值为

Answer 1

由余弦定理得：cosA=（b²+c²-a²)/2bc cosB=(a²+c²-b²)/2ac cosC=(a²+b²-c²)/2ab
代入原式=（b²+c²-a²)/2+(a²+c²-b²)/2+(a²+b²-c²)/2
=(a²+b²+c²)/2

Answer 2

用余弦定理
bccosA+cacosB+abcosC
=bc(b�0�5+c�0�5-a�0�5)/2bc+ca(c�0�5+a�0�5-b�0�5)/2ca+ab(a�0�5+b�0�5-c�0�5)/2ab
=(b�0�5+c�0�5-a�0�5+c�0�5+a�0�5-b�0�5+a�0�5+b�0�5-c�0�5)/2
=(a�0�5+b�0�5+c�0�5)/2
=(9+16+36)/2
=61/2

Answer 3

61/2