Question 890243
<pre>
{{{drawing(400,800/3,-1,11,-1,7,

triangle(0,0,3sqrt(3),3,0,6),
triangle(0,6,0,0,10.39230485,0),
arc(0,0,2.5,-2.5,30,90),
arc(0,6,2.5,-2.5,270,330),locate(1,.5,"30°"),
locate(2.5,5,y), locate(-.3,3,6),
arc(3sqrt(3),3,2.5,-2.5,150,210), locate(8.7,.5,"z°")
locate(-.2,0,A),locate(10.4,0,B), locate(5.2,3.5,D),locate(0,6.5,C) 
 )}}} 

Since the three angles of &#916;ACD have equal measure,
&#916;ACD is equilateral, and &#8736;CAD = &#8736;CDA = &#8736;C = 60°  

Therefore AC = 6 = AD = CD = y.

So y = 6

Since &#8736;C = &#8736;CAD = &#8736;CDA = 60°, &#8736;CAB = &#8736;CAD + &#8736;DAB = 60° + 30° = 90°

So &#916;ABC is a right triangle, and &#8736;B and &#8736;C are complementary,

So &#8736;B = 90°-&#8736;C = 90°-60° = 30° = z°.

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We could have gotten z° from the fact that &#916;ABD is isosceles,
since AD = 6 and BD = 6, and the base angles are equal, so 
&#8736;B = &#8736;DAB = 30° = z².

Edwin</pre>