Question 1193135
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Let's sketch it and see:

{{{drawing(400,4228/17,-1,16,-1,9.57,
locate(.8,1.3,59^o), 
locate(8.5,8.5,121^o),
locate(0,0,A), locate(15,0,B), locate(5,9.3,D), locate(9.8,9.3,C),
line(0,0,5.15,8.57), line(5.15,8.57,15-5.15,8.57),line(15-5.15,8.57, 15,0),line(15,0,0,0) )}}}

Draw DE perpendicular to parallel bases AB and DC.

{{{drawing(400,4228/17,-1,16,-1,9.57,
locate(.8,1.3,59^o), green(line(5.15,0,5.15,8.57)),
locate(8.5,8.5,121^o),locate(5.15,0,E),
locate(0,0,A), locate(15,0,B), locate(5,9.3,D), locate(9.8,9.3,C),
line(0,0,5.15,8.57), line(5.15,8.57,15-5.15,8.57),line(15-5.15,8.57, 15,0),line(15,0,0,0) )}}}

1. Now since triangle AEB is a right triangle, you can find m∠ADE.
And you know that m∠EDC = 90<sup>o</sup>, now you can find the measure
of the whole angle at D, namely m∠ADC, which could be called m∠D 
before we drew in DE.

2. Now that you have the measures of all but one of the interior angles, 
use the fact that the sum of the measures of the interior angles of
any quadrilateral is 360<sup>o</sup> to find m∠B.

Edwin</pre>