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We need help with this question. Sorry but it’s taken from a French math book, so we had to translate the it.ABC is a triangle that has: ACB = 86° and ABC = 37°.The bisector of angle ACBIntersects segment [AB] at M.Calculate angles BMC and AMCThis is what we have done so far:BCM-MBC=(86:2)-37=43-37So, 180-6=174BMC = 180 - 6 = 174

Sagot :

We want to calculate angles BMC and AMC, as you can observe, these angles are supplementary.

Thus using triangle relations, we start by calculating angle BMC.

We can agree that;

[tex]\begin{gathered} \angle CBM+\angle MCB+\angle BMC=180 \\ \angle CBM=37 \\ \angle MCB=\frac{86}{2}=43 \\ \angle BMC=180-43-37=100^o \end{gathered}[/tex]

And;

[tex]\begin{gathered} \angle BMC+\angle AMC=180 \\ 100+\angle AMC=180 \\ \angle AMC=80^o \end{gathered}[/tex]

Thus;

[tex]\begin{gathered} \angle AMC=80^0 \\ \angle BMC=100^o \end{gathered}[/tex]

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