help with geometry

BMC + A = 201

BMC + CMA = 180

BMC + C also = 201

 

3BMC + CMA + C +A = 582, and we know CMA + C + A = 180, so 3BMC = 582-180 = 402, and thus BMC = 134

 

B = MCB, and finish the last step to get B

Let's see

angle BMC +angle A = 201

 

consider triangle CMA

2

2A=201-A

3A=201

A=67 degrees

so

angle CMA= 180-2*67 = 46degrees

 

so angle B = half of 46 = 23 degrees.

 

I have not carefully checked this.