AM is the median of triangle ABC. Find the angle ACB (in degrees) of the triangle if MAB = 60∘, AMC = 120∘.

We put all the data from the condition of the problem on the drawing:

The median AM bisects the BC side. Let’s denote this by serifs on CM and MB.

Note that the AMC Angle is 120 degrees.
We also make a note of 60 degrees in the MAB Corner.

Let’s take a closer look at the triangle ABM
The AMB angle is adjacent to the AMC angle, which means:

Angle AMB = 180 degrees – Angle AMC = 180 – 120 = 60 degrees.

In turn, the sum of the angles of any triangle is 180 degrees, which means that we can find the angle ABM:

Angle ABM = 180 degrees – Angle AMB – Angle MAB = 180 – 60 – 60 = 60 degrees.

Since all 3 angles are equal to 60 degrees, the ABM triangle is equilateral, that is:

AM = AB = MB.

To find the value of the angle ACB, consider the triangle AFM
We found out that AM = MB, therefore, AM = MC. It turns out that the ACM triangle is isosceles.

By the property of an isosceles triangle: the angles at the base are equal, that is,

Angle AFM = Angle CAM.

Since the sum of the angles of the triangle is 180 degrees, it turns out:

ACM + CAM + AMC = 180 degrees.

Since CAM = ACM, we get the equation:

ACM + ACM + 120 = 180;

2 * ACM = 180 – 120 = 60;

ACM = 60/2 = 30 degrees.

Since ACB = ACM, the sought Angle ACB is 30 degrees.

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