we know that
The measurement of the external angle is the semi-difference of the arcs it comprises.
so
in this problem
m∠VWX=[tex]\frac{1}{2} (arc\ UY-arc\ VX)[/tex]
we have
m∠VWX=[tex]43\°[/tex]
[tex]arc\ UY=125\°[/tex]
Substitute and solve for arc VX
[tex]43\°=\frac{1}{2}*(125\°-arc\ VX)[/tex]
[tex]86\°=(125\°-arc\ VX)[/tex]
[tex]arc\ VX=125\°-86\°=39\°[/tex]
therefore
the answer is
the measure of arc VX is [tex]39\°[/tex]
