Constructing an equilateral spherical triangle on the surface of the earth How would you go about this when given the latitude and longitude of two of the vertices. I would assume the longitude of the third vertex would have to bisect the arc that connects the two known vertices. But I can't figure out how to get the latitude of the unknown vertex other than "guess and check"

The cosine of the distance between the two given points is given by the Spherical Law of Cosines:
$$
\cos(\delta)=\sin(\beta_1)\sin(\beta_2)+\cos(\beta_1)\cos(\beta_2)\cos(\lambda_2-\lambda_1)\tag1
$$
where $\beta_k$ are the latitudes and $\lambda_k$ are the longitudes of the two given points, $p_k$.
The equal angles of the triangle can be computed also using the Spherical Law of Cosines:
$$
\begin{align}
\cos(\Delta)
&=\frac{\cos(\delta)-\cos^2(\delta)}{\sin^2(\delta)}\\
&=\frac{\cos(\delta)}{1+\cos(\delta)}\tag2
\end{align}
$$
Note that $\lim\limits_{\delta\to0}\Delta=\frac\pi3$, as in the planar case.
Now we can compute $\mathrm{B}_2$, the azimuth of $p_2$ as viewed from $p_1$.
$$
\begin{align}
\cos(\mathrm{B}_2)&=\frac{\sin(\beta_2)-\sin(\beta_1)\cos(\delta)}{\cos(\beta_1)\sin(\delta)}\tag{Law of Cosines}\\
\sin(\mathrm{B}_2)&=\frac{\cos(\beta_2)\sin(\lambda_2-\lambda_1)}{\sin(\delta)}\tag{Law of Sines}\\
\tan\left(\frac{\mathrm{B}_2}2\right)&=\frac{\cos(\beta_1)\cos(\beta_2)\sin(\lambda_2-\lambda_1)}{\sin(\delta-\beta_1)+\sin(\beta_2)}\tag3
\end{align}
$$
We can compute the latitude of the two possible third points
$$
\sin(\beta_3)=\sin(\beta_1)\cos(\delta)+\cos(\beta_1)\sin(\delta)\cos(\mathrm{B_2}\pm\Delta)\tag4
$$
Then we can compute the longitude
$$
\begin{align}
\cos(\lambda_3-\lambda_1)&=\frac{\cos(\delta)-\sin(\beta_1)\sin(\beta_3)}{\cos(\beta_1)\cos(\beta_3)}\tag{Law of Cosines}\\
\sin(\lambda_3-\lambda_1)&=\frac{\sin(\delta)\sin(\mathrm{B_2}\pm\Delta)}{\cos(\beta_3)}\tag{Law of Sines}\\
\tan\left(\frac{\lambda_3-\lambda_1}2\right)&=\frac{\cos(\beta_1)\sin(\delta)\sin(\mathrm{B_2}\pm\Delta)}{\cos(\beta_1+\beta_3)+\cos(\delta)}\tag{5}
\end{align}
$$
