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"
 A: 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
$$
\begin{align}
\cos(\lambda_3-\lambda_1)&=\frac{\cos(\delta)-\sin(\beta_1)\sin(\beta_3)}{\cos(\beta_1)\cos(\beta_3)}\tag{Law of Cosines}\\
\sin(\lambda_3-\lambda_1)&=\frac{\sin(\delta)\sin(\mathrm{B_2}\pm\Delta)}{\cos(\beta_3)}\tag{Law of Sines}\\
\tan\left(\frac{\lambda_3-\lambda_1}2\right)&=\frac{\cos(\beta_1)\sin(\delta)\sin(\mathrm{B_2}\pm\Delta)}{\cos(\beta_1+\beta_3)+\cos(\delta)}\tag{5}
\end{align}
$$
