Finding out the value of $\angle DQC$ in a trapezium $ABQD$ where $\angle DCB$ = 30$^\circ$ In this below diagram, $\angle ABC=60^\circ, \angle DCB=30^\circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$? 

At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$\frac{1}{2}(AD+BC)$ ×$AP$ = $\frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$ 
$BC$ = $PQ$ ..................              (1)
And now, we can show their perimeter are equal with the below equation: 
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$ 
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)

I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$. 
By trigonometry and from the traingle $APB$, we can write that:-
$\frac{AB}{AP}$ = $\mathrm{cosec} 60^\circ$
$AB$ = $\frac{2AP}{\sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$\frac{DC}{DE}$ = $\mathrm{cosec} 30^\circ$
$DC$ = 2$DE$ 
$CD$ = 2$DE$..........(4)
Let us denote the $\angle DQE = $$\theta$
So, now from the equation (2), we can get:-
$2AP$ + $\frac{2AP}{\sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$\frac{2\sqrt 3AP + 2AP - \sqrt 3AP}{\sqrt 3} $ = $DQ$
$AP × \frac{2+ \sqrt 3}{\sqrt 3} $ = $DQ$
$\frac{2+ \sqrt 3}{\sqrt 3} $ = $\frac{DQ}{AP} $
$\frac{DQ}{DE} $ = $\frac{2+ \sqrt 3}{\sqrt 3} $......($AP = DE$  according to the diagram)
$\frac{DE}{DQ} $ = $\frac{\sqrt 3}{2 +\sqrt 3} $
$\sin \theta$ = $\frac{\sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $\sin^\text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation. 
(5) is invalid because the real value of $\sin \theta$ is above 1 which is impossible in this case. For this reason, the value of $\theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
 A: The problem has no solution. Let's assume that there is one. Then we notice that increasing or decreasing the length of $AD$ will not change the fact that areas/ perimeters are the same. Indeed, if we increase the length by $x$, the perimeter of both trapezoids will increase by $2x$, and the areas will both increase by $x\cdot AE$. Then let's do $x=-|AD|$. Then in your picture $A=D$ and $P=E$. I will therefore drop any reference to $D$ and $E$ from now on.
$ABC$ is a right angle triangle, with the acute angles $30^\circ$ and $60^\circ$. Then $$AC=\frac{\sqrt 3}2 BC$$ and $$AB=\frac 12 BC$$
In the right angle triangle $ABC$ I can write the area in two ways to get $$BC\cdot AP=AB\cdot AC=BC^2\frac{\sqrt 3}{4}$$
Therefore $AP=BC \frac{\sqrt 3}{4}$. You already showed that $PQ=BC$. So in the right angle triangle $APQ$ you can write $$\tan\theta=\frac{AP}{PQ}=\frac{\sqrt 3}4$$
Now lets see that this does not verify equal perimeter requirement: $$AB+AC+BC=BC\left(\frac 12+\frac{\sqrt 3}2+1\right)$$
$$AP+PQ+QA=BC\left(\frac{\sqrt 3}{4}+1+\sqrt{1^2+\left(\frac{\sqrt 3}{4}\right)^2}\right)$$
You can see that there is something with $\sqrt{19}$ in the second equation, that it's not there in the first.
Note: you could get to the same conclusion even if you explicitly carry around $AD\ne 0$.
A: 
\begin{align}
   \operatorname{perimeter}(APQD) &= \operatorname{perimeter}(ABCD) \\
   (3+\sqrt 3)x + 2y + CQ+DQ &= (6+2\sqrt 3)x + 2y \\
   CQ +DQ &= (3+\sqrt 3)x
\end{align}
\begin{align}
   \operatorname{area}(APQD) &= \operatorname{area}(ABCD) \\
   \frac 12(3x+2y+CQ)(\sqrt 3x) &= \frac 12(4x+2y)((\sqrt 3x)) \\
   CQ &= x \\
\hline
   DQ &= (2+\sqrt 3)x \\
   (DQ)^2 &= (7+4\sqrt 3)x^2
\end{align}
But, then, $(7+4\sqrt 3)x^2=(DQ)^2 = (\sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
