Question 1036184
<pre><b>
This figure is drawn to scale:

{{{drawing(400,52400/233,-10,223,-10,121,
locate(100,0,AB=213), locate(214,56,111),
line(0,0,213,0),line(213,0,213,111), line(213,111,28,111),
line(28,111,0,0), line(28,111,28,0), line(0,0,213,111),
red(locate(7,17,theta),locate(37,13,phi),
arc(28,1036/71,20,-20,270,388),
arc(0,0,50,-50,27,76)), rectangle(200,0,213,13),
rectangle(213,111,200,98),locate(14,0,28),
rectangle(28,111,41,98), locate(29,25,F),
locate(0,0,A), locate(213,0,B), locate(210,121,C),
locate(28,0,E),locate(25,121,D),locate(120,121,185) 

 )}}}
<pre>
AE = AB-EB = AB-DC = 213-185 = 28

tan(&#8736;DAE) = {{{"opposite"/"adjacent"}}} = {{{DE/"AE"}}} = {{{CB/"AE"}}} = {{{111/28}}}

&#8736;DAE = 75.84236735°

tan(&#8736;CAB) = {{{"opposite"/"adjacent"}}} = {{{CB/AB}}} = {{{111/213}}}

&#8736;CAB = 27.52522574°

&#8736;<font face="symbol">q</font> = &#8736;DAE-&#8736;CAB = 75.84236735°-27.52522574° = 48.31714161°

------------------
&#8736;<font face="symbol">f</font> is an exterior angle to &#916;AEF,
so it is the sum of the two remote interior angles of &#916;AEF,

so &#8736;<font face="symbol">f</font> = &#8736;CAE + &#8736;AEF = 27.52522574°+ 90° = 117.5252257°

Edwin</pre></b>