Given a triangle ABC. Point M lies on side BC. It is known that AB = BM and AM = MC, the angle B is equal to 120 °

| May 26, 2021 | education

Given a triangle ABC. Point M lies on side BC. It is known that AB = BM and AM = MC, the angle B is equal to 120 °. Find the rest of the corners of triangle ABC.

Consider a triangle ABM, it is isosceles (AB = BM, by hypothesis). In an isosceles triangle, the angles at the base are equal, we find them.
Angle BAM = Angle BMA = (180 ° – Angle B) / 2 = (180 ° – 120 °) / 2 = 30 °.
Now consider the triangle AMC, it is isosceles (AM = MC, by condition). Find the angle AMC at the apex of this isosceles triangle. It is adjacent to BMA, which means:
Angle AMC = 180 ° – Angle BMA = 180 ° – 30 ° = 150 °.
Knowing the angle at the apex, we find the angles at the base of the isosceles triangle AMC:
Angle MAC = MCA = (180 ° – 150 °) / 2 = 15 °.
Angle BAC = BAM + MAC = 30 ° + 15 ° = 45 °
Answer: angle A 45 °, angle C 15 °.



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