∠A+∠C=2∠B ∠B =60

tan(A+C)=(tanA+tanC)/( 1-tanA tanC)= tanB

tanA+tanC = tanB( 1-tanA tan)=√3( 1-2-√3)=-√3-3

tanAtanC=2+√3

Through the formula x 2+(p+q) x+pq = (x+p) (x+q)

Let a 2-(√ 3+3) a+(2+√ 3) = 0.

(a- 1)(a-2-√3)= 0a = 1a = 2+√3

TanA= 1 tanC=2+√3, that is, A = 45 C = 180-60-45 = 75.

AB high CD through C. Then BCD is a right triangle of 30, 60 and 90.

ACD is a right triangle of 45, 45 and 90.

Then BD = 2 √ 3, CD = 6 = AD.

AC=6√2

S=(6+2√3)*6/2= 18+6√3