设三角形ABC的内角ABC的对边长分别为abc 已知a方-c方=2b 且sin=4cosAsinC.求b

Question

如题所述

Answer 1

解析：∵sinB=4cosAsinC
∴sinB/sinC=b/c=4cosA
∵cosA=(b^2+c^2-a^2)/2bc,
∴b/c=4(b^2+c^2-a^2)/2bc,
b/c=2(b^2-2b)/bc,
b^2=2b^2-4b,
b^2-4b=0,
∵b≠0,∴b=4