The diagram will be somewhat like this :
<BAC= <ACB=<CBA=60°
cosθ = da and sinθ = √(a2-d2)d
cos(120°-θ) = 1-da = 1a - cosθ
Expanding cos(120°-θ) we get,
cosθ2 + √3 sinθ2 = 1a
Substituting the values of cosθ and sinθ we get ,
√(3a2-d2) = 2-d
or a = 2√(d2-d+13)
Which is the required answer.