已知三角形ＡＢＣ的三边长分别是2m+3,m2+2m,m2+3m+3,则最大角的度数是多少?越详细越好

Question

已知三角形ＡＢＣ的三边长分别是2m+3,m2+2m,m2+3m+3,则最大角的度数是多少?
越详细越好

Answer 1

三角形三边均为正
2m+3>0 m>-3/2
m²+2m>0 m(m+2)>0 m>0或m0,m可取任意实数.
综上,得m>0
m²+3m+3-(m²+2m)=m+3>0
m²+3m+3-2m-3=m²+m>0
m²+3m+3为最长边,所对的角为最大角.
由余弦定理得
cos(最大角)=[(2m+3)²+(m²+2m)²-(m²+3m+3)²]/[2(2m+3)(m²+2m)]
=[(2m+3)²+(m²+2m+m²+3m+3)(m²+2m-m²-3m-3)]/[2(2m+3)(m²+2m)]
=[(2m+3)²+(2m²+5m+3)(-m-3)]/[2(2m+3)(m²+2m)]
=[(2m+3)²-(m+1)(2m+3)(m+3)]/[2(2m+3)(m²+2m)]
=(2m+3)[(2m+3)-(m+1)(m+3)]/[2(2m+3)(m²+2m)]
=(2m+3)(2m+3-m²-4m-3)/[2(2m+3)(m²+2m)]
=(2m+3)(-2m-m²)/[2(2m+3)(m²+2m)]
=-(m²+2m)(2m+3)/[2(2m+3)(m²+2m)]
=-1/2
最大角=120°