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The problem is as follows:

The figure from below shows a figure. Find the requested angle $x$ using the information given.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&120^{\circ}\\ 2.&130^{\circ}\\ 3.&100^{\circ}\\ 4.&110^{\circ}\\ 5.&105^{\circ}\\ \end{array}$

How exactly can I find the requested angle?. So far the only thing which I was able to spot is that $\angle POM= \frac{\theta}{2}$.

Then this would meant that:

$\triangle MOR$ to be isosceles. But I don't know exactly what else can it be found.

The thing is that I was able to conclude that the angle $\angle POM = \frac{\theta}{2}$ to be that way by tracing a bisector line $PO$.

Then there's an identity relating the angle formed by the bisector lines in a triangle one interior bisector and the other an exterior bisector. But again, can someone help me here?. Please include a diagram or drawing in the answer. Since reading from solely algebraic expressions don't help much in this kinds of figures.

Chris Steinbeck Bell
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2 Answers2

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The answer should be $120$.

The first thing you should do is draw the diagram properly ( the figure given is just not good) , then observe that by the angle conditions given to you we get $QR=MR=PM=QM \implies \Delta MQR$ is an equilateral triangle . Then we get $\angle \theta= 30$ , and since $QN$ is the angle bisector $\implies \angle QNM=90 \implies \angle x= 30+90= 120$ .

enter image description here

Sunaina Pati
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  • Does drawing and exterior circle contributed in something to the solution of this problem?. Because I don't see anything that relates it to the solution. Or is it a construction made on purpose?. In other words was the circle necessary? In similar approaches I got confused if one of the points do lie in the circle, in this case O did not lie in the circle.I don't know how does it help to solve the problem knowing that $QN$ is bisector?. I think if $MO$ is bisector the it makes sense that way. Can you specify and answer my doubts please? – Chris Steinbeck Bell Oct 28 '20 at 00:35
  • I didn't get you properly , by exterior circle did you mean circle PQR ? well yeah it wasn't necessary for the problem . I drew it for just just to share more info , like in this diagram we have PQR a right-angle triangle.. and also PQR is a 30-90-60 triangle , with M as it's circumcentre . All this are additional infos you can get from the problem . But drawing circle wasn't necessary – Sunaina Pati Oct 28 '20 at 01:32
  • @ChrisSteinbeckBell regarding your second doubt . Did you understand that $\Delta OMR $ is an equilateral triangle . Now, after this it well known that in an equilateral triangle , the angle bisectors are also the altitudes ( and also the median , but it isn't necessary here ) . This actually follows from congruence , which is quite easy , and you can try to prove that on your own ( if you havn't) , and if you get stuck feel free to ask . :) – Sunaina Pati Oct 28 '20 at 01:38
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$$\angle MPQ = \angle MQP = \theta \leadsto PM = MQ$$

$$\angle RMQ = \angle RQM = 2\theta \leadsto MR = RQ$$

Hence $MQ= PM = MR = RQ$, and thus $\triangle MQR$ is equilateral.

This leads to $\theta = 30^\circ$. By observing exterior angles:

$$x = \angle OMN + \angle QNM = \theta + (180^\circ - \theta - 2\theta) = 120^\circ$$

player3236
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