Question

在三角形ABC中,角A，B，C的对边分别为a,b,c,向量m=(2cosc/2,-sinc),n(cosc/2,2sinc),且m⊥n,若a^2=2b^2+

若a^2=2b^2+c^2,求tanA的值

Answer 1

m⊥n
=> m.n=0
(2cosc/2,-sinc).(cosc/2,2sinc)=0
2(cosc/2)^2-2(sinc)^2=0
cosC +1 - 2(1-(cosC)^2 ) =0
2(cosC)^2 + cosC-1=0
(2cosC-1)(cosC+1)=0
C = π/3
a^2=2b^2+c^2
= b^2 + c^2 +b^2
by cosine rule
b^2 = -2bccosA
cosA = -b/(2c)

B= (2π/3 -A)
sinB = (√3/2)cosA + (1/2)sinA
(1/2)tanA + (√3/2) = sinB/cosA
tanA = 2sinB/cosA - √3
= 2(bsinC/c)/cosA - √3
= 2(b[√3/2]/c)/[-b/(2c)] - √3
= -2√3-√3
= - 3√3

追问

答案写成（-3根号147）/7，对吗？