[Math] Finding distance from the center of gravity of an equilateral triangle to a point outside.

Question

For a triangulation project, I would like to know if it's possible to find the distance from the center of gravity G of an equilateral triangle to a point R outside of it.

The only information we know is the side length d of the triangle, and the angles made by lines going through R and each vertex A, B, C of the triangle. Is there a formula for this?

Answer 1

Let say P is nearer to BC of the equilateral $\triangle ABC$ with sides $= d$. If $G$ is the centroid, then $BG = h$, a known quantity.

In $\triangle APB$, $\theta$ is known because $\alpha$ and $\beta$ are known. Applying sine law to it, we can find $PB$.

In $\triangle BGP$, $GP$ can be found by applying cosine law to it.

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In case “the angles made by lines going through R and each vertex A, B, C of the triangle” means $\theta$ and $\rho$ are known instead. See the second figure.

Apply sine law to $\triangle ACR$, we have $\dfrac {AR}{\sin (60^0 + \lambda)} = \dfrac {d}{\sin \rho}$

Similarly, from $\triangle ABR$, we have $\dfrac {AR}{\sin (60^0 + \beta)} = \dfrac {d}{\sin \theta}$

Cancelling AP and d from the two equations, we get

$\dfrac {\sin (60^0 + \beta)}{\sin (60^0 + \lambda)} = \dfrac {\sin \theta}{\sin \rho}$

$\dfrac {\sin (60^0 + \beta)}{\sin (60^0 + 180^0 - \beta - \theta - \rho)} = \dfrac {\sin \theta}{\sin \rho}$

$\dfrac {\sin (60^0 + \beta)}{\sin (\beta + \theta + \rho – 60^0)} = \dfrac {\sin \theta}{\sin \rho}$

The last equation contains only one unknown ($\beta$) should be solvable. Once it is known, $\alpha$ is also known. This is exactly case-1.