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In the diagram below, $AM=BM=CM$ and $\angle BMC+\angle A = 201^\circ.$ Find $\angle B$ in degrees.

 Feb 12, 2021

Best Answer 

 #1
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Refer to image below:

180-BMC = CMA                  BAC = CMA          MCB= CBM

 

180-BMC + A + A = 180                   BMC+A=201    SO    A = 201-BMC

180-BMC     +  2* (201-BMC) = 180

∴  BMC = 134o

                                          (   THEN   CMA = 180 - 134 = 46            AND      A and  C = 67 )

 

Since BMC = 134    then the other two angle in the triangle on the left =   (180-134)/2 = 23o = angle B

 

 Feb 12, 2021
 #1
avatar
+1
Best Answer

Refer to image below:

180-BMC = CMA                  BAC = CMA          MCB= CBM

 

180-BMC + A + A = 180                   BMC+A=201    SO    A = 201-BMC

180-BMC     +  2* (201-BMC) = 180

∴  BMC = 134o

                                          (   THEN   CMA = 180 - 134 = 46            AND      A and  C = 67 )

 

Since BMC = 134    then the other two angle in the triangle on the left =   (180-134)/2 = 23o = angle B

 

Guest Feb 12, 2021

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