Question

在三角形ABC中,内角A,B,C的对边分别为a,b,c,已知a²-c²=2b,且sinB=4cosAsinC,求b

Answer 1

由sinB=4cosAsinC得sinB/sinC=4cosA
由正弦定理 余弦定理得
所以b/c=2（b2+c2-a2）/bc
由上式得b2-2（a2-c2）=0
又因为a2-c2=2b
所以b2-4b=0
解得b=4或0
因为b＞0
所以b=4

Answer 2

sinB=4cosAsinC
sinB/sinC = 4cosA
b/c = 4cosA
by cosine rule
a^2= b^2+c^2-2bccosA
=b^2+c^2- (1/2)b^2
a^2-c^2 = (1/2)b^2 (1)
a^2-c^2=2b (2)

from (1) and (2)
(1/2)b^2=2b
b(b-4) =0
b=4