Suppose I am standing at point $A$ on Earth (a 3D sphere of a known radius) and suppose $B$ is another point a few miles away. The longitude and latitude of points $A$ and $B$ are $(\lambda_1,\varphi_1)$ and $(\lambda_2,\varphi_2)$ respectively. What is the angle from $A$ to $B$? In other words, If I was holding a compass whose needle points north while standing at point $A$, at what angle will point $B$ be?
[Math] Angle between two points on a sphere
anglespherical trigonometry
Related Solutions
$\DeclareMathOperator{\proj}{proj}$This can be handled easily using vector algebra. Let $A$ and $B$ be the specified points on the unit sphere, viewed as vectors in space, and assume $A \neq \pm B$ and $\cos r < A \cdot B$ ($A$ is outside the circle $C$ of radius $r$ centered at $B$). Let $U_{1}$ be the unit vector orthogonal to $B$ and tangent to the "short" great circle arc from $B$ to $A$, and let $U_{2} = B \times U_{1}$.
Theorem: If $$ \theta = \arccos\left[\frac{(A \cdot B)\tan r}{A \cdot U_{1}}\right], $$ then the points $T_{1}$ and $T_{2}$ are $$ (\cos r)B + \sin r\bigl((\cos\theta) U_{1} \pm (\sin\theta) U_{2}\bigr). $$
Computational note:
- $U_{1}$ is the second vector obtained from the Gram-Schmidt algorithm on the ordered basis $\{B, A\}$; that is, $$ U_{1} = \frac{A - \proj_{B}A}{\|A - \proj_{B}A\|}. $$
Proof of Theorem: Every point on the circle $C$ has the form $$ T = (\cos r)B + \sin r\bigl((\cos\theta) U_{1} \pm (\sin\theta) U_{2}\bigr) \tag{1} $$ for some real $\theta$.
If $T$ is an arbitrary point of $C$, let $A' = A - T$ and $B' = B - T$ denote the displacements to $A$ and $B$. The projections of these vectors to the tangent plane at $T$ are $$ A'' = A' - (A' \cdot T)T,\qquad B'' = B' - (B' \cdot T)T, $$ and $T$ is a point of tangency if and only if $A'' \perp B''$. Expanding the dot product, using the fact that $T$ is a unit vector $(T \cdot T = 1$) lying on the circle of radius $r$ centered at $B$ ($B \cdot T = \cos r$), gives (omitting most of the algebra) $$ 0 = A'' \cdot B'' = A' \cdot B' - (A' \cdot T)(B' \cdot T) = A \cdot B - (\cos r)A \cdot T, $$ or $$ A \cdot T = \frac{A \cdot B}{\cos r}. \tag{2} $$ Substituting (1) in (2), using the fact that $A \cdot U_{2} = 0$, and solving for $\theta$ gives the value in the theorem.
Indeed both the formulae are identical. This can be shown in the following manner.
In the spherical polar coordinate system $\theta$ polar angle is the angle between the zenith direction and a line segment OP. But $\theta$ is nothing but the co latitude. Since we do know that latitude
$\theta$ = 90 - colatitude and also $\cos(90 - \theta)$ = $\sin\theta$. Substituting this into the original identity reveals that the identities are equal.
Best Answer
There are two laws of cosines for spherical triangles: $$\begin{align}\cos a&=\cos b\cos c+\sin b\sin c\cos A\\ \cos A&=-\cos B\cos C+\sin B\sin C\cos a\end{align}$$ Where $A$, $B$, and $C$ are the angles of the spherical triangle and $a$, $b$, and $c$ are the opposite side. Here you know one angle and two sides of the spherical triangle with vertices at the cities and the north pole: $$\begin{align}b=\frac{\pi}2-\phi_1,&&c=\frac{\pi}2-\phi_2,&&\text{and}\quad A=\lambda_2-\lambda_1\end{align}$$ So you can apply the first of the two laws to get $$\cos a=\sin\phi_1\sin\phi_2+\cos\phi_1\cos\phi_2\cos(\lambda_2-\lambda_1)$$ Then the distance between the two cities is $Ra$, where $R$ is the radius of the earth and $a$ is the angle we just computed in radians. Oh, but you wanted the bearing. There is also a law of sines: $$\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$$ So we can now get $$\sin C=\frac{\sin A\sin c}{\sin a}$$ And $C$ is the bearing from city $1$ to city $2$ measured east of north.