Question 396756
Let's draw it to scale. Here's the trapezoid:

{{{drawing(400,400,-.5,.5,-.05,.95,

line(-cos(67*pi/180),0,cos(67*pi/180),0),
line(-cos(67*pi/180),0,-.220941202,.4),

line(cos(67*pi/180),0,.220941202,.4),



line(-.220941202,.4,.220941202,.4) )}}}

and here is the rest of the isosceles triangle, of
which the isosceles trapezoid is a part:

{{{drawing(400,400,-.5,.5,-.05,.95,

line(-cos(67*pi/180),0,cos(67*pi/180),0),
line(-cos(67*pi/180),0,-.220941202,.4),

line(cos(67*pi/180),0,.220941202,.4),
locate(-.016,.8,"46°"),
triangle(-cos(67*pi/180),0,cos(67*pi/180),0,0,sin(67*pi/180)),

line(-.220941202,.4,.220941202,.4) )}}}


Next we draw the midsegment, which splits the figure in
half.  It splits the 46° vertex angle into two 23° angles,

and the green line is perpendicular to both bases of
the trapezoid:

{{{drawing(400,400,-.5,.5,-.05,.95,
rectangle(-.04,0,0,.04), rectangle(-.04,.4,0,.44),
line(-cos(67*pi/180),0,cos(67*pi/180),0),
line(-cos(67*pi/180),0,-.220941202,.4),

line(cos(67*pi/180),0,.220941202,.4),
locate(-.06,.77,"23°"),
triangle(-cos(67*pi/180),0,cos(67*pi/180),0,0,sin(67*pi/180)),
green(line(0,0,0,sin(67*pi/180))),
line(-.220941202,.4,.220941202,.4) )}}}


Now we'll just look at the left half, which is a right triangle,
actually two right triangles:

{{{drawing(400,400,-.5,.5,-.05,.95,
rectangle(-.04,0,0,.04), rectangle(-.04,.4,0,.44),

line(-cos(67*pi/180),0,0,0),
line(-cos(67*pi/180),0,-.220941202,.4),

locate(-.06,.77,"23°"),
triangle(-cos(67*pi/180),0,0,0,0,sin(67*pi/180)),
green(line(0,0,0,sin(67*pi/180))),
line(-.220941202,.4,0,.4) )}}}

Since the two acute angles of a right triangle are complementary,
the other acute angle in both right triangles is 

90°-23° or 67°:

{{{drawing(400,400,-.5,.5,-.05,.95,
rectangle(-.04,0,0,.04), rectangle(-.04,.4,0,.44),

line(-cos(67*pi/180),0,0,0),
line(-cos(67*pi/180),0,-.220941202,.4),
locate(-.35,.05,"67°"),locate(-.18,.45,"67°"),

locate(-.06,.77,"23°"),
triangle(-cos(67*pi/180),0,0,0,0,sin(67*pi/180)),
green(line(0,0,0,sin(67*pi/180))),
line(-.220941202,.4,0,.4) )}}}

and the angle supplementary to the upper 67° angle is 180°-67° = 113°

{{{drawing(400,400,-.5,.5,-.05,.95,
rectangle(-.04,0,0,.04), rectangle(-.04,.4,0,.44),

line(-cos(67*pi/180),0,0,0),
line(-cos(67*pi/180),0,-.220941202,.4),
locate(-.35,.05,"67°"),locate(-.18,.45,"67°"),

locate(-.06,.77,"23°"), locate(-.21,.37,"113°"),
triangle(-cos(67*pi/180),0,0,0,0,sin(67*pi/180)),
green(line(0,0,0,sin(67*pi/180))),
line(-.220941202,.4,0,.4) )}}}

So therefore the original isosceles trapezoid has an acute
angle of 67° and an obtuse angle of 113°: 

{{{drawing(400,400,-.5,.5,-.05,.95,

line(-cos(67*pi/180),0,cos(67*pi/180),0),
line(-cos(67*pi/180),0,-.220941202,.4),

line(cos(67*pi/180),0,.220941202,.4),
locate(-.21,.37,"113°"), locate(-.35,.05,"67°"),



line(-.220941202,.4,.220941202,.4) )}}}