Sides & Angles

Area = $\frac {1}{2} \times Base \times Height$

thus, 3 = $\frac {1}{2} \times d \times h$ ...(i) {where h is height of triangle}

$\tan (\theta) = \frac {height}{base}$

thus, $\frac {3}{2} = \frac {h}{p}$ ...(ii) [where p is distance of foot of perpendicular (dropped from A on BC) from B]

And, $\frac {1}{2} = \frac {h}{d-p}$ ...(iii)

from (ii) & (iii) eliminate p, to get, $h = \frac{3d}{8}$ ... (iv)

now use (iv) in (i),

$3 = \frac{1} {2} \times { d }^{ 2 } \times \frac {3}{8}$

thus finally, we get answer, $\boxed { d\quad =\quad 4 }$

ALTER:

Move this triangle to cartesian plane & follow simple coordinate geometry. (with BC along positive x - axis.) use: slope = $\tan \theta$ where $\theta$ is angle with positive x-axis.