Question

bsinA＝√2a－acosB，求B

Answer 1

2-2accos60 9=4a^2+a^2-2a^2 a=根号3 c=2a=2根号3 所以, a=根号3, c=2根号3 bsina=√3acosb sinBsinA=√3sinAcosB sinB=。

Answer 2

bsinA =√2a－acosB
b/(√2-cosB) = a/sinA
=>
√2-cosB = sinB
(sinB+cosB)=√2
1+sin2B =2
sin2B =1/2
2B = π/6 or 5π/6 (rej)
B =π/12
ie
B =π/12

Answer 3

bsinA =√2a－acosB
b/(√2-cosB) = a/sinA
=>
√2-cosB = sinB
(sinB+cosB)=√2
1+sin2B =2
sin2B =1/2
2B = π/6 or 5π/6 (rej)
B =π/12
ie
B =π/12