Let y=AB=CD (y represents the lengths of AB and CD); let a=BD
Cosine Rule:
132=a2+y2-2aycos(3x-4),
162=a2+y2-2aycos(47).
Subtract these two equations:
162-132=87=2ay(cos(3x-4)-cos(47)).
cos(3x-4)=87/(2ay)+cos(47).
Since the RHS must be greater than cos(47), cos(3x-4)>cos(47). This means 3x-4<47°, 3x<51°, x<17° because, for example, if x=15°, 3x-4=41° and cos(41)>cos(47)
This reduces the possible values of x to 1 or 15. We have just shown that x=15° is a possibility:
cos(41)-cos(47)=87/(2ay)=0.0727 approx. So 2ay=87/0.0727=1196.514 approx, so ay=598.257.
When x=1°, 3x-4=-1°, which is a negative angle which cannot apply to an angle in a triangle. Therefore, the solution to this question is x=15°, even though we can't determine the length of BD; or the equal lengths AB and CD.
Note that x=15° is a possible solution, but not an actual solution. The actual solutions differ for different values of a and y (different lengths for BD and AB, CD). In fact, x=15° produces no solutions for the lengths of BD, AB, CD. Below is a specific example of the quadrilateral in which a=y=AB=BD=CD in length (triangles ABD and BDC are isosceles). (In this scaled figure x=13.936° approx.)