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.