Question 1108438
Here is a sketch of the parallelogram, with sides of length {{{x}}} and {{{y}}} ,
and a diagonal of length 16.5.
The diagonal divides the parallelogram into two triangles
whose angles measure {{{red("36 ° 10 ' ")}}} , {{{red("14°30' ")}}} and {{{red(A)}}} .
 
{{{drawing(600,200,-1,17,-1,5,
triangle(0,0,15.97,4.13,12.59,0),
triangle(0,0,15.97,4.13,3.39,4.13),
red(arc(0,0,4,4,-14.5,0)),
red(arc(15.97,4.13,4.5,4.5,129.33,165.5)),
red(arc(12.59,0,1.5,1.5,180,309.33)),
locate(8,2.1,16.5),locate(2.1,0.8,red(14.5^o)),
locate(14.1,3.9,red(36^o)),locate(14.8,3.7,red("10 '")),
locate(12.3,0.6,red(A)),locate(6.2,0,x),
locate(13.9,1.7,y)
)}}}
Applying Law of Sines to the triangle with the angles and sides labeled, we get

{{{16.5/sin(red(A))}}}{{{"="}}}{{{x/sin(red("36 ° 10 ' "))}}}{{{"="}}}{{{y/sin(red("14°30' "))}}}
{{{red(A)=180^o-(red("36 ° 10 ' ")+red("14°30' "))=180^o-50^o}}}{{{"40'"=129^o}}}{{{"20'"}}}
A calculator tells us that accurate to 6 decimal places: 
{{{sin(red(A))}}}{{{"="}}}{{{"sin ("}}}{{{129^o}}}{{{"20' )"}}}{{{"="}}}{{{0.773472}}} ,
{{{"sin ("}}}{{{36^o}}}{{{"10' )"}}}{{{"="}}}{{{0.590136}}} and
{{{"sin ("}}}{{{14^o}}}{{{"30' )"}}}{{{"="}}}{{{0.250380}}} .
Substituting those values,
{{{16.5/0.773472}}}{{{"="}}}{{{x/0.590136}}}{{{"="}}}{{{y/0.250380}}}
Solving the equations {{{16.5/0.773472}}}{{{"="}}}{{{x/0.590136}}} and {{{16.5/0.773472}}}{{{"="}}}{{{y/0.250380}}} ,
we get values for {{{x}}} and {{{y}}} .
Calculated to 2 decimal places, the lengths are:
{{{x}}}{{{"="}}}{{{16.5*0.590136/0.773472}}}{{{"="}}}{{{12.59}}} and
{{{y}}}{{{"="}}}{{{16.5*0.250380/0.773472}}}{{{"="}}}{{{5.34}}} .