Question 1210494: The centroid of triangle ABC is G. Find x, in degrees.
Found 2 solutions by CPhill, n2:
Answer by CPhill(2103)   (Show Source):
You can put this solution on YOUR website! The value of $x$, which is $\angle AGB$, is $\mathbf{120^\circ}$.
Here is the reasoning based on the properties of an equilateral triangle and its centroid.
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## Centroid and Medians in an Equilateral Triangle
1. **Medians are also Angle Bisectors and Altitudes:** In an **equilateral triangle** ($\triangle ABC$), the medians (which intersect at the centroid $G$) are also the angle bisectors and the altitudes.
2. **Symmetry:** Because $\triangle ABC$ is equilateral, it has three-fold rotational symmetry. This means that the area, angles, and distances formed by the centroid to the three vertices are identical.
3. **Angles around the Centroid:** The centroid $G$ divides the triangle into three smaller triangles: $\triangle AGB$, $\triangle BGC$, and $\triangle CGA$. The angles formed at the centroid ($\angle AGB$, $\angle BGC$, and $\angle CGA$) must sum to $360^\circ$.
$$\angle AGB + \angle BGC + \angle CGA = 360^\circ$$
4. **Equal Angles:** Due to the symmetry of the equilateral triangle, these three central angles must be equal:
$$\angle AGB = \angle BGC = \angle CGA$$
### Calculation
Let $x = \angle AGB$.
$$x + x + x = 360^\circ$$
$$3x = 360^\circ$$
$$x = \frac{360^\circ}{3} = \mathbf{120^\circ}$$
Therefore, $\angle AGB = 120^\circ$.
Answer by n2(19)  (Show Source):
You can put this solution on YOUR website! .
The problem's formulation is incomplete and can not be considered as a true/proper Math problem.
Is not a subject for discussion or solution.
Is the subject to throw to a garbage bin.
It's hooliganism to post defective tasks on the forum.
The fact that @CPhill is using artificial intelligence to solve this problem is the attempt
to deceive a reader and tells about his irresponsible attitude towards his duties.
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