Question 1204459
<pre>

{{{drawing(800,320,-1.5,13,-1.5,4.3,
locate(7.5,2.9,114^o), locate(3.8,1.9,8.6), locate(6,0,x),
line(0,0,11.11075603,0), line(11.11075603,0,8.028791668,3.081964366),
line(8.028791668,3.081964366,0,0),locate(1,.5,21^o) )}}}

For enrichment purposes, here's an alternate way to do the problem.  It's a
little longer, but you could do it this way if you forgot the law of sines.
Draw the altitude h from the vertex of the 114<sup>o</sup> angle. That splits the 114<sup>o</sup> angle
into the complement of 21<sup>o</sup> or 90<sup>o</sup>-21<sup>o</sup>=69<sup>o</sup>, and 114<sup>o</sup>-69<sup>o</sup>=45<sup>o</sup>. It also splits x, making

x = p+q.

{{{drawing(800,320,-1.5,13,-1.5,4.3,
red(line(8.028791668,3.081964366,8.028791668,0)),
locate(4,0,p), locate(9.4,0,q), red(locate(8.13,1.5,h)),

locate(7.5,2.9,69^o), locate(8.1,2.7,45^o),locate(3.8,1.9,8.6),
line(0,0,11.11075603,0), line(11.11075603,0,8.028791668,3.081964366),
line(8.028791668,3.081964366,0,0),locate(1,.5,21^o) )}}}

Then using cosine = adjacent/hypotenuse in the rt. triangle on the left,
{{{p/8.6=cos(21^o)}}}
{{{p=8.6*cos(21^o)}}}
{{{p=8.028791668}}}

Then using sine = opposite/hypotenuse in the same rt. triangle,
{{{h/8.6=sin(21^o)}}}
{{{h=8.6*sin(21^o)}}}
{{{h=3.081964366}}}

Since the rt. triangle on the right has a 45<sup>o</sup> angle, it is 
isosceles, so q = h = 3.081964366.  Then

x = p + q = 8.028791668 + 3.081964366 = 11.11075603.

x rounds to 11.1

Edwin</pre>