Triangle ABC is isosceles, AB = BC, angle B = 100 degrees, angle MAB = 10 degrees

Triangle ABC is isosceles, AB = BC, angle B = 100 degrees, angle MAB = 10 degrees, angle MBA = 20 degrees. Find the angle of the BMC.

Let’s determine the value of the angle AMB. Angle АМВ = (180 – МАВ – МВА) = (180 – 10 – 20) = 150.

Determine the value of the angle MВС. Angle MВС = ABC – MВA = 110 – 20 = 90.

Let the sought angle BMC = X0, then the angle BCM = (90 – X) 0.

In triangle ABM we apply the theorem of sines.

AB / Sin150 = BM / Sin10.

AB = BM * Sin150 / Sin10.

In the IUD triangle, apply the sine theorem.

ВС / SinX = BM / Sin (90-X) = BM / CosX.

BC = BM * SinX / CosX = BM * tgX.

Triangle ABC is isosceles, AB = BC, then:

BM * tgX = BM * Sin150 / Sin10.

tgX = Sin150 / Sin10 = 0.5 / 0.173 = 2.88.

ВМС angle = arctg2.88 ≈ 71.

Answer: The angle of the ВМС is 71.