Question 1210567
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In cyclic quadrilateral PQRS, angle P = 30, angle Q = 60, PQ = 4, QR = 8.
Find the largest side in quadrilateral PQRS, in degrees.
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        Obviously,  the question in the post is posed in mathematically illiterate way.

        But I will not pay attention to it.


        My goal here is to prove that such a quadrilateral,  as described in the post, 

        DOES  NOT  EXIST  and  CAN  NOT  EXIST.



<pre>
Let's consider triangle PRQ, formed by sides PQ and QR and the diagonal PR of quadrilateral PQRS.  

Since the quadrilateral PQRS is cyclic, it is convex, so the diagonal PR lies inside this quadrilateral.


In this triangle, we are given the side lengths  PQ = 4  and  QR = 8.
We also are given the contained angle  Q  of  60°.


So, we can compute the opposite side PR of this triangle using the cosine law

    PR = {{{sqrt(4^2 + 8^2 - 2*4*8*cos(60^o))}}} = {{{sqrt(80-64*(1/2))}}} = {{{sqrt(80-32)}}} = {{{sqrt(48)}}} = {{{4*sqrt(3)}}}.


Now we see that  {{{PQ^2}}} + {{{PR^2}}} = {{{4^2}}} + {{{48}}} = 16 + 48 = 64 = {{{8^2}}} = {{{QR^2}}}.

Hence, triangle PQR is a right-angled triangle with the legs PQ and PR with the right angle at vertex P.


But angle P is 30°, as it is given in the problem - - - so, it CAN NOT contain angle QPR of 90°.


This CONTRADICTION proves that quadrilateral PQRS as described in the post, DOES NOT EXIST and CAN NOT EXIST.


        The problem describes something geometric shape that does not exist and can not exist,
        and the problem tries to implant this wrong/false idea into the reader's mind.
</pre>

Thus the problem is refuted, &nbsp;i.e. &nbsp;killed to the death and ruined into dust.


The posed problem tries to conceive a reader, &nbsp;as it do many other problems from the same source,
which I call &nbsp;" the Laboratory for false defective problems in &nbsp;Geometry ".