# Problem involving angles of a triangle [closed]

$\triangle ABC$ has $AC=BC$ and $\angle ACB = 106^\circ$ . $M$ is a point in $\triangle ABC$ such that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$ .What is the measure( in degree ) of $\angle BMC$ ?

## closed as off-topic by S.C.B., user91500, JonMark Perry, TastyRomeo, TheGeekGreekFeb 16 '17 at 9:07

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• Hello, and welcome to MSE @Sophia. Could you show us what you have tried? We would appreciate some general context and additional information for this problem. – S.C.B. Feb 15 '17 at 14:43
• It is a simple angle chasing, once you notice that $\widehat{CAB}=\widehat{CBA}=37^\circ$. – Jack D'Aurizio Feb 15 '17 at 14:47
• yes , It is .But I still can't solve . :( – SophaVisa Khun Feb 15 '17 at 14:53
• @JackD'Aurizio I don't think it's quite simple angle calcs - there will be some construction involved which I can't immediately see, or trigonometry (use of sine rule or similar). – Joffan Feb 15 '17 at 15:16
• Yes , I used sinus law . It helps but I still stuck . – SophaVisa Khun Feb 15 '17 at 15:21

Draw the orthogonal bisector $h$ of edge $AB$. Then since $AC = BC$, point $C$ lies on $h$. Choose point $D$ on $h$ such that $\angle \, DBA = 60^{\circ}$ and both $C$ and $D$ lie on the same side of $AB$. Then $BD = AD$, hence triangle $ABD$ is equilateral. After angle chasing, $\angle \, BAM = 30^{\circ}$ so $\angle \, DAM = 30^{\circ}$.
Thus $\angle \, DAC = \angle \, DAM - \angle \, MAC = 30^{\circ}-7^{\circ} = 23^{\circ}$. Consequently, $$\angle \, DAC = 23^{\circ} = \angle \, MCA$$ so $MC$ is parallel to $AD$. Therefore the quad $AMCD$ is trapezoid and $\angle \, DAM = 30^{\circ} = \angle \, ADC$. Hence $AMCD$ is an equilateral trapezoid with $AM = DC$. That is why the orthogonal bisector $l$ of $AD$ is also the orthogonal bisector of $MC$. Since $B$ lies on $l$ due to the fact that triangle $ABD$ is equilateral, $BM = MC$ and hence $$\angle \, BMC = \angle \, BCM = \angle \, ACB - \angle \, MCA = 106^{\circ} - 23^{\circ} = 83^{\circ}$$