Question

在三角形ABC中,已知b=(根号3-1)a,C=30度,求角A与角B的度数

Answer 1

=(√3-1)a

sinB = (√3-1)sinA

C=30°

A +B=150°

sinB = (√3-1)sin(150°-B) = (√3-1)(sin150°cosB-cos150°sinB) = (√3-1)(1/2cosB+√3/2sinB)

2sinB = (√3-1)cosB+(3-√3)sinB

(√3-1)sinB = (√3-1)cosB

tanB=1

B=45°

A=180°-B-C=105°