Question 1037714
An isosceles trapezoid looks like this:
{{{drawing(300,250,-1,8,-1,6.5,
line(0,0,7,0),line(2,5.5,5,5.5),
line(0,0,2,5.5),line(7,0,5,5.5),
green(arc(0,0,2,2,-70,0)),green(arc(7,0,2,2,-180,-110)),
red(arc(2,5.5,2,2,0,110)),red(arc(2,5.5,1.5,1.5,0,110)),
red(arc(5,5.5,2,2,70,180)),red(arc(5,5.5,1.5,1.5,70,180))
)}}} , with two congruent {{{green(acute)}}} angles and two congruent {{{red(obtuse)}}} angles.
The {{{green(acute)}}} angles are supplementary to the {{{red(obtuse)}}} angles.
The vertices in a trapezoid (or any polygon) are named going around in order (clockwise or counterclockwise).
No matter how the vertices in your isosceles trapezoid are named,
{{{A}}} and {{{C}}}angles are opposite vertices/angles.
The opposite angles in an isosceles trapezoid
(one acute and one obtuse) are supplementary,
meaning that their measures add up to {{{180^o}}} , so
{{{A+C=180^0}}} , meaning that {{{14x+(3x+10)=180}}} .
Now, all you have to do is solve that equation:
{{{14x+(3x+10)=180}}}
{{{(14x+3x)+10=180}}}
{{{17x+10=180}}}
{{{17x=180-10}}}
{{{17x=170}}}
{{{x=170/17}}}
{{{highlight(x=10)}}} 
If we are curious, we can figure out the measures of the angles as
{{{C=14x^o=14*(10^o)=140^o}}} and
{{{A=(3x+10)^o=(3*10+10)^o=(30+10)^o=40^o}}} .
So, the isosceles trapezoid in the problem looks like this:
{{{drawing(800,250,-0.5,7.5,-0.5,2,
line(0,0,7,0),line(2,1.68,5,1.68),
line(0,0,2,1.68),line(7,0,5,1.68),
green(arc(0,0,1.3,1.3,-40,0)),green(arc(7,0,1.3,1.3,-180,-140)),
red(arc(2,1.68,1,1,0,140)),red(arc(2,1.68,0.75,0.75,0,140)),
red(arc(5,1.68,1,1,40,180)),red(arc(5,1.68,0.75,0.75,40,180)),
locate(2,1.68,red(140^o)),locate(0.3,0.3,green(40^o))
)}}} 
 
NOTE:
A rectangle is considered a special case of a trapezoid,
and it can be considered an isosceles trapezoid.
In that case, all four angles would measure {{{90^o}}} ,
with no acute or obtuse angle,
but opposite angles would still be supplementary.
That is not the case with the problem above,
but the reasoning above would have worked just the same,
even if we had figured out that both angles were right angles.