Constructing an equilateral spherical triangle on the surface of the earth

Question

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"

Answer 1

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