+129852 Angle BCD is also 30.3 So..BDC = 90-30.3 = 59.7
So we can find BC with the Law of Sines
BC/sin59.7 = 67.7 / sin 90
BC = 67.7* sin 59.7 = 58.5 cm
So AC - BC = AB = 67.7 - 58.5 = 9.2 cm
Notice that this doesn't seem to square with the picture as presented...the probable reason is that "theta" seems much greater than 30.3 degrees......However, we can only work with what we're given....
+1006 One property of a rhombus is that its opposite angles are the same. Thus,
cos(30.3) = b/67.7
67.7(cos(30.3)) = b
$${\mathtt{67.7}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30.3}}^\circ\right)} = {\mathtt{58.451\: \!878\: \!776\: \!093\: \!9}}$$
58.45 cm is the length of segment AB. Rounded, that equals 58.5 cm, which is what CPhill got, the difference being I didn't get points 
.
+118677
<EDC=180-30.3 = 149.7 degrees Cointerior angels on parallel lines.
So
You can use Sin(59.7) etc to get BC
AB = AC-BC
Not so difficult. Everything is easy when you know how. 
Actually,
It would have been easier if i had said <AED=<ACD (opposite angles in a rhombus are equal.)
and gone from there.
+129852 Angle BCD is also 30.3 So..BDC = 90-30.3 = 59.7
So we can find BC with the Law of Sines
BC/sin59.7 = 67.7 / sin 90
BC = 67.7* sin 59.7 = 58.5 cm
So AC - BC = AB = 67.7 - 58.5 = 9.2 cm
Notice that this doesn't seem to square with the picture as presented...the probable reason is that "theta" seems much greater than 30.3 degrees......However, we can only work with what we're given....
