Help to understanding the problem

Question

The problem:

Three poles standing at the points A,B and C subtend angles $\alpha$, $\beta$ and $\gamma$ respectively,at the circumcentre of the triangle ABC.If the heights of these poles are in arithmetic progression; then show that $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression.

Now,what I could not understand is subtending of the angle part,precisely how a point subtend angle at another point? So,what I am looking for a proper explanation of the problem statement with a figure,since it's troubling me from sometime.

PS:I am not looking for the solution (as of now) or any hints regarding the solution,just a clear explanation will be appreciated.

---

My solution using Moron's interpretation,

Let $a$ $b$ and $c$ are the length of three sides of the poles and $O$ be the circumcentre then, $$ \cot \alpha = \frac{OA}{a}$$ $$ \cot \beta = \frac{OB}{b}$$

$$ \cot \gamma = \frac{OC}{c}$$

As O is the circumcentre,$OA = OB = OC = k $(say)

Again, $a$ $b$ and $c$ are in arithmetic progression,hence,

$$2 \cdot \frac{k}{\cot \alpha} = \frac{k}{\cot \beta} + \frac{k}{\cot \gamma}$$

Canceling $k$ from both sides,

$$2 \cdot \frac{1}{\cot \alpha} = \frac{1}{\cot \beta} + \frac{1}{\cot \gamma}$$

Hence, $\cot \alpha$, $\cot \beta$ and $\cot \gamma $ are in harmonic progression. (QED)

Answer 1

My guess is the poles are of (different) heights $h_A$, $h_B$, $h_C$ and the angle is from the foot of pole to the circumcentre of $\triangle ABC$ to the top of the pole.

Answer 2

I would read it that the poles have a diameter>0. The pole at A has diameter so the angle seen from the circumcenter is $\alpha$. See if that works.