Question

在△ABC中，b²=ac,c=2a，则cosB=？

Answer 1

cosB=3/4
解：三角形ABC中
cosB=(a^2+c^2-b^2)/(2ac)
b^2=ac
c=2a
则b^2=2a^2
c^2=4a^2带入上式有
cosB
=(a^2+4a^2-2a^2)/(2*2a^2)
=3a^2/4a^2
=3/4

Answer 2

根据余弦定理：
cosB=(a²+c²-b²)/2ac
=(a²+4a²-2a²)/(4a²)
=3/4

Answer 3

b²=ac=a*(2a)=2a²
cosB
= (a²+c²-b²) / (2ac)
= [a²+(2a)²-2a²] / (2a*2a)
=(a²+4a²-2a²) / (4a²)
=3a²/(4a²)
=3/4