# 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.

• There is no solution. You cannot have both perimeters and areas the same. – Andrei Jan 24 at 7:42
• That's why I got stucked. I often get this kind of faulty question but find no satisfied answer. The source of the problem is also enigmatic. Thank you for telling me that point. I frankly didn't know that point. Should I delete this post? Otherwise anyone will be confused. – Anirban Niloy Jan 24 at 7:48
• I did not know it either. But it's easy to prove. I will post that as an answer. – Andrei Jan 24 at 7:51
• Okay. I will check that out later. I'll be eagerly waiting. – Anirban Niloy Jan 24 at 7:53

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 $$AD\ne 0$$.
• $AP\perp BC$. For the other one, if you have in a triangle an angle of $60^\circ$ and an angle of $30^\circ$, the last one has to be $90^\circ$ – Andrei Jan 24 at 8:44
• But here we can see in that diagram that $\angle DCE = 30^\circ$. Steven gregory solved the value of $\angle DQC but how can we justify that? – Anirban Niloy Jan 24 at 8:46 • Make$A=D$. In the picture in the other answer use$y=0– Andrei Jan 24 at 8:48 • Oops, sorry. You told me in your answer. It's a great honour that you gave your precious time to me. Tnx for your help and support. – Anirban Niloy Jan 24 at 8:50 \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. • In triangleDEQ$you have$EQ=EC+CQ=4x$. Then$DQ^2=DE^2+EQ^2=3x^2+16x^2=19x^2$but$x\sqrt{19}\ne(2+\sqrt 3)x$– Andrei Jan 24 at 8:52 • @stevengregory If you use the tangent function,$\tan\angle DQC=\frac{\sqrt 3 x}{4x}$you get a different value for the angle, only about$23.4^\circ$– Andrei Jan 24 at 9:15 • Yeah, you're right. If we consider the condition as a right one (including area and perimeter are equal of both the field mentioned in the question), then$27.65^\circ\approx23.4^\circ\$. That makes no sense. – Anirban Niloy Jan 24 at 10:20