Question

在△ABC中，角A.B.C的对边分别为a.b.c且cos2C+3cosC=1.c=√7.

Answer 1

cos2C+3cosC=1
2cos²C-1+3cosC=1
2cos²C+3cosC-2=0
(cosC+2)(2cosC-1)=0
cosC+2＞0
∴2cosC-1=0
∴cosC=1/2
∴C=60°

正弦定理：
a/sinA=b/sinB=c/sinC
a=csinA/sin60°=√7sinA/(√3/2) = 2√(7/3)sinA
b=2√(7/3)sinB

面积S=1/2absinC = 3√3/2
ab=3√3/sinC = 3√3/(√3/2) = 6

余弦定理：
c²=a²+b²-2abcosC=a²+b²+ab=(a+b)²-ab = 7
{ 2√(7/3)sinA+2√(7/3)sinB }² - 6 = 7
28/3{sinA+sinB }² =13
{sinA+sinB }² =39/28
sinA+sinB＞0
sinA+sinB =√(39/28) = √273/14

Answer 2

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