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

QuestionMachine Feb 12, 2021

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Best Answer

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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 = 134^o

( 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 = 23^o = angle B

Guest Feb 12, 2021

1⁺⁰ Answers

+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 = 134^o

( 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 = 23^o = angle B

Guest Feb 12, 2021

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