问题
解答题
设△ABC的内角A,B,C的对边分别为a,b,c,且A=60°,c=3b.求: (Ⅰ)
(Ⅱ)cotB+cot C的值. |
答案
(Ⅰ)由余弦定理得a2=b2+c2-2bccosA=(
c)2+c2-2•1 3
c•c•1 3
=1 2
c2.7 9
∴
=a c
.7 3
(Ⅱ)cotB+cotC=
=cosBsinC+cosCsinB sinBsinC
=sin(B+C) sinBsinC
,sinA sinBsinC
由正弦定理和(Ⅰ)的结论得
=sinA sinBsinC
•1 sinA
=a2 bc
•2 3
=
c27 9
c⋅c1 3
=14 3 3
.14 3 9
故cotB+cotC=
.14 3 9