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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+0 Answers

#1
+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