In the adjoining figure, $$AB$$ || $$DC$$. Also, $$PG=PF$$. The measure of angle $$x$$ in degrees is

Given that AB || CD,

$$ \angle BCD = 180\degree - \angle ABC = 180\degree- 142\degree = 38\degree$$ (Interior angles are supplementary)

$$\angle CEJ = \angle BCD = 38\degree$$ (Vertically opposite angles)

$$\angle EPJ = 360\degree - 90\degree- 90\degree - \angle CEJ = 360\degree - 90\degree-90\degree - 38\degree =142\degree$$ (Angle sum property of a quadrilateral)

Since PG = PF, $$\triangle PGF$$ is an isosceles triangle  and $$ \angle PGF = \angle PFG$$

$$\angle PGF = \dfrac{1}{2} (180\degree - 142\degree) = 19\degree$$

$$ x = 180\degree-\angle PGF = 180\degree - 19\degree = 161\degree$$ (Straight angle)