已知在三角形ABC中,角ABC的对边分别为abc.且4cos^2*C/2-cos2C=7/2.a+b=5.c=根号7 求角C的大小和三角形的面积
问题描述：
已知在三角形ABC中,角ABC的对边分别为abc.且4cos^2*C/2-cos2C=7/2.a+b=5.c=根号7 求角C的大小和三角形的面积
答
∵ab≤(a+b)^2/4
=25/4
ab最大值 25/4＜9
CosC
=(a^2+b^2-c^2)/2ab
=[(a+b)^2-c^2-2ab]/2ab
=(25-7-2ab)/2ab
=(18-2ab)2ab
=9/ab-1＞0 (9/25/4=36/25＞1)
∴0°＜C＜90°
∵4cos^2C/2-cos2C
=2（2cos^2C/2-1）+2-cos2C
=2cosC+2-cos2C
=2cosC+2-2cos^2C+1
=-2cos^2C+2cosC+3
=7/2
∴cos^2C-cosC+1/4=0
(cosC-1/2) ^2=0
CosC=±1/2
C=60°
C=120°舍掉(∵0°＜C＜90°上边已确定)
SinC=sin60°=√3/2
∵.a+b=5
∴(a+b) ^2
=a^2+b^2+2ab
=25
A^2+b^2=25-2ab
c ^2=√7 ^2=7
C^2=a^2+b^2-2abcosC
ab=(a^2+b^2-c^2)/2cosC
=(25-2ab-7)/2*1/2
=18-2ab
3ab=18
ab=6
S=1/2absinC
=1/2*6*√3/2
=3√3/2
C=60° S=3√3/2
吉林 汪清LLX