Question

在三角形ABC中,角ABC的对边分别为abc,cosC=4/5,c=2bcosA

1.求证：A=B
2.若b=5,求c的值
3.求cos(4A-π/6)的值
急求

Answer 1

(1)c=2bcosA
则 c^2=2bccosA 根据余弦定理：
a^2=b^2+c^2-2bccosA
则a^2=b^2+c^2-c^2=b^2，a=b
∴△ABC是等腰三角形 A=B
(2) ∵cosC=4/5>0 ∴C为锐角
则sinC=√(1-(cosC)^2)=3/5
∵S△ABC=1/2absinC=5/2
∴ab=25/3
c^2=a^2+b^2-2abcosC
=a^2+b^2-2ab+2ab-2abcosC
=(a-b)2+2ab(1-cosC)
=0+2*25/3*(1-4/5)
=10/3
c=√30/3