the sum of all interior angles in a polygon is
180(n-2), where n = sides, well, this is a QUADrilateral, so it has 4 sides, so the total is 360°.
now, let's find what angle C is first,
[tex]\bf \stackrel{A}{(2x-40)}+\stackrel{B}{(116)}+C+\stackrel{D}{(x)}=360\implies C+3x+76=360
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C+3x=284\implies C=284-3x[/tex]
now, recall the "inscribed quadrilateral conjecture", where opposite angles are "supplementary angles", thus
[tex]\bf \stackrel{\measuredangle A}{(2x-40)}+\stackrel{\measuredangle C}{(284-3x)}=180\implies -x+244=180
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64=x\\\\
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\measuredangle A=2(64)-40[/tex]
