∴2B=A+C,A+B+C= 180,
∴B=60,A=60 -θ,C=60 +θ
And ∵ 1/SINA, 3 √ 2/2 SINB, 1/SINC became arithmetic progression.
∴3√2/sinb= 1/sina+ 1/sinc
2√6 = 1/sin(60-θ)+ 1/sin(60+θ)
cosθ/[(cosθ)^2- 1/4]=√2
The solution is cosθ=√2/2 or -√2/4 (abbreviated).
∴θ=45
(2) sinA/a=sinB/b=sinC/c can be obtained by sine theorem,
A=(√6-√2)
Sina = sin (60-45) = (√ 6-√ 2)/4,sinb = sin60 = √ 3/2,sinc = sin (60+θ) = (√ 6+√ 2)/4,b = 2 √ 3。
∴s= 1/2absinc= 1/2*(√6-√2)*2√3*(√6+√2)=√3