Question 206765
{{{drawing(300,300,0,8,0,8,
   line(2,2,2,6),
   line(2,2,6,2),
   line(2,6,3.9,6.7),
   line(3.9,6.7,6,2),
   red( line(2,6,6,2) ),

   line(2,2.5,2.5,2.5),
   line(2.5,2,2.5,2.5),

   line(3.47,6.52,3.67,6.16),
   line(4.09,6.31,3.67,6.16),

   locate(1.7,1.9,D),
   locate(5.8,1.9,C),
   locate(1.7,6.2,A),
   locate(3.75,7.2,B)
) }}}
Since AD=CD and angle D is a right angle, Triangle ADC is an isosceles right triangle so angles DAC and  DCA are 45 degrees.
 
The tangent of angle BAC (by SOH-CAH-TOA) is {{{opposite/adjacent = BC/BA = 5.7/3 = 1.9}}}
so the measure of angle BAC is arctan(1.9) = 62.2    On your calculator, arctan or inverse tangent is on the same button as tan, and looks like tan to the -1 power (which it isn't).

So the angle at A is the sum of the angles DAC and BAC which is 45 + 62.2 = 107.2 degrees.