请问各位大神,这道题怎么做?

2025-01-06 07:09:24
推荐回答(1个)
回答1:

1)
cosA = ( b^+c^-a^ )/ 2bc = (c-b)/2b
cosB = ( a^+c^-b^ )/ 2ac = (b+c)/2a

cos2B = 2(cosB)^-1 = (c-b)/2b

也即:cosA = cos2B

又 a^=b^+ bc , 所以a大于b 2B小于180度

A和2B都小于180度,只能是A=2B
2)3B^2= B^2+BC --> 2B=C --> B = C / 2 --->...