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The problem is as follows:

The figure from below shows two triangles. Assume $PR=QS$. Using this information find $x$.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&20^{\circ}\\ 2.&18^{\circ}\\ 3.&22^{\circ}\\ 4.&24^{\circ}\\ 4.&30^{\circ}\\ \end{array}$

What I attempted to do to solve this problem was to use the identity which relates the exterior sum of the two angles in a triangle. In other words:

$\angle PQS= 6x$

But that's it, I came stuck there. Since I found this problem in a section belonging to triangle congruency, I think it should be solved using such approach. Can someone help me here?. If so, what sort of the cases related with congruency of triangles is it?. Could it be that is side-angle-side?.

Please include a drawing in the solution as I dont know how to spot those in triangles.

Chris Steinbeck Bell
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2 Answers2

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enter image description here

Construct point $A$ (as shown in the picture) such that $\overline{PS}=\overline{RA}$. Note that $$\overline{QS}=\overline{PR}=\overline{PS}+\overline{SR}=\overline{RA}+\overline{SR}=\overline{SA}$$ Therefore $$\angle SQA=\angle SAQ\implies 2x+\angle RQA=\angle RAQ\quad\quad(1)$$ Also note that $$\angle RQA+\angle RAQ=4x\quad\quad(2)$$

Solve system of equations and we have $$\angle RQA=x,\quad\angle RAQ=3x$$

Now since $\angle SPQ=\angle RAQ=3x$, we have $\overline{QP}=\overline{QA}$. Hence we can say that $$\triangle PQS\cong\triangle AQR\quad(S.A.S.)$$ so $$\overline{QS}=\overline{QR}\implies \angle QSR=\angle QRS=4x$$

Finally $$4x+4x+2x=180^{\circ}\implies \color{red}{x=18^{\circ}}$$

  • That's beautiful! I know this may be hard to answer, but how did you get the idea for that construction of point A? – Ameet Sharma Nov 21 '20 at 16:13
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    To be honest I am not exactly sure. Since $\overline{QS}=\overline{PR}$ but point $S$ is not an endpoint of $\overline{PR}$ (we don't like that), I constructed point $A$ to get an isosceles, and saw if there's anything can be done after that. Fortunately, there was, and I eventually solved this really quick. –  Nov 21 '20 at 17:28
  • @Student1058 You're awesome that's a neat explanation just as helpful as one can get. To be honest, I also thought that such sort of construction would be intended. But I failed to spot the congruency of the two triangles and the identities you had used. It seems that the catch here was to build an isosceles triangle thus this can be solved. Perhaps if you have time can you offer a hand with this one? – Chris Steinbeck Bell Nov 22 '20 at 02:54
  • @ChrisSteinbeckBell Thanks and I will surely check this one out. –  Nov 22 '20 at 03:13
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enter image description here[![enter image description here][2]][2]

This figure satisfies the conditions. If $\angle QPR=\angle PQR=54^o$ then $\angle SQR=36^o$ then $ PR=QR=QS$. So correct option is $18^o$. Other options are wrong.

Moreover we can solve following equation:

$2x+4x+3x+\alpha=180$

Where $\alpha=\angle PQS$

Among numerous solutions only one set of solution; $(x, \alpha)=(18, 18)$ satisfies the condition

sirous
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