Find the distance between Helsinki and Seattle How do I find the distance between Helsinki and Seattle along the shortest route? This is actually a question from a mathematically-based Astronomy book. I am not in a science or math class. I'm just trying to learn Astronomy on my own and this is one of the questions in the book.
Here are the Questions: Find the distance between Helsinki and Seattle along the shortest route. Where is the northernmost point of the route, and what is its distance from the North Pole? The longitude of Helsinki is $25$ degrees E and latitude $60$ degrees; the longitude of Seattle is $122$ degrees W and latitude $48$ degrees. Assume that the radius of the Earth is $6{,}370$ km(kilometers).
The answers the book gives are: 1) $7{,}640$ kilometers, 2) the Northernmost point is $79$ degrees N, $45$ degrees W, and 3) in North Greenland, $1{,}250$ km from the North Pole. 
I know that the central angle can be found using the Great Circle equation:
$$Δσ = \arccos( \sin φ_1 × \sin φ_2 + \cos φ_1 × \cos φ_2 × \cos Δλ ),$$
where $Δσ$ is the central angle,
$φ_1$ and $λ_1$ are the latitude and longitude of the first city,
and $φ_2$ and $λ_2$ are the latitude and longitude of the second city. But I don't know how to find $Δλ$ and the other two answers (where the northernmost point of the route is, and what its distance from the North Pole is).
\begin{align}
φ_1 &= 60° \\
λ_1 &= 25° \\
φ_2 &= 48° \\
λ_2 &= -122°
\end{align}
Plug in:
\begin{align}
Δσ &= 68.72° \\
Δσ &= 1.199 \text{ radians}
\end{align}
$$s = rΔσ = 7640 \text{ km}$$
 A: So I think the question is for two points given in spherical coordinates $\vec{r}_1=(R,\theta_1,\phi_1)$ and $\vec{r}_2=(R,\theta_2,\phi_2)$, how to find the shortest distance between them on the spherical surface with radius $R$?
First we need to find the angle between $\vec{r}_1$ and $\vec{r}_2$, $\alpha$, which can be easily obtained by dot product.
$$\hat{r}_1=\sin\theta_1\cos\phi_1 \hat{x}+\sin\theta_1\sin\phi_1\hat{y}+\cos\theta_1\hat{z}$$
$$\hat{r}_2=\sin\theta_2\cos\phi_2 \hat{x}+\sin\theta_2\sin\phi_2\hat{y}+\cos\theta_2\hat{z}$$
$$\hat{r}_1\cdot\hat{r}_2=\sin_1\theta_1\sin\theta_2\cos\phi_1\cos\phi_2+\sin\theta_1\sin\theta_2\sin\phi_1\sin\phi_2+\cos\theta_1\cos\theta_2$$
$$=\sin\theta_1\sin\theta_2\cos\left(\phi_1-\phi_2\right)+\cos\theta_1\cos\theta_2$$
$$\alpha=\cos^{-1}\left[\sin\theta_1\sin\theta_2\cos\left(\phi_1-\phi_2\right)+\cos\theta_1\cos\theta_2\right]$$
And therefore the distance is
$$D=R\alpha=R\cos^{-1}\left[\sin\theta_1\sin\theta_2\cos\left(\phi_1-\phi_2\right)+\cos\theta_1\cos\theta_2\right]$$
A: For 1), you can use the spherical law of cosines which can be reduced to
$$ D = R \arccos{(\sin \lambda_A \sin \lambda_B + \cos \lambda_A \cos \lambda_B \cos(L_A - L_B))} $$
where $(\lambda_A, L_A)$ are the latitude and longitude of point A and similarly for B.
For 3) once you know 2), and denoting the point as C, you can just use
$$ D' = R (90 - \lambda_C) \frac{\pi}{180}$$
where $\lambda_C$ is the latitude of C - the expression is just the angle from the North Pole in radians. This is a much simpler application of the same law:
$$ D' = R \arccos(\cos(90 - \lambda_C)) $$
2) may be harder - I don't know how to do it off the top of my head. I guess the easiest way is to get the cross product as a comment suggests. The angle between that and the North Pole is the latitude of C.
A: 
Key: P = north pole, A = Seattle, B = Helsinki, V = northernmost point on great circle route.
Reference: Wikipedia article on spherical trigonometry
As pointed out in the OP, the great circle distance can be found by use of the spherical low of cosines for sides.  The convention in spherical trigonometry is that co-latitude is used as the measure of the angle from the north pole to a point, so the co-latitude is zero at the north pole and 90 degrees at the equator.  The co-latitudes of Seattle and Helsinki are $42^{\circ}$ and $30^{\circ}$ respectively.
Just for the record, the calculation of the great circle distance from Seattle to Helsinki goes like this:
$$\begin{align}
\cos AB &= \cos a \; \cos b + \sin a \; \sin b \; \cos P \\
&= \cos 42^{\circ} \cos 30^{\circ} + \sin 42^{\circ} \sin 30^{\circ} \cos 147^{\circ} \\
&= 0.362992
\end{align}$$
So $AB = 1.1993 \text{ radians} = 68.7^{\circ}$, and the distance is $1.1993 \times 6370 = 7640$ km.
Next we compute the angle $A$, again using the spherical law of cosines.
$$\cos a = \cos b \; \cos AB + \sin b \; \sin AB \; \cos A$$
so
$$\begin{align}
\cos A &= \frac{\cos a - \cos b \; \cos AB}{\sin b \; \sin AB} \\
&= \frac{\cos 42^{\circ} - \cos 30^{\circ} \cos 68.7^{\circ}} {\sin 30^{\circ} \sin 68.7^{\circ}} \\
&= 0.9200
\end{align}$$ so $A = 23.0^{\circ}$
The remaining calculations have to do with the laws of spherical right triangles, since angle $AVP = 90^{\circ}$.
$$\begin{align}
\sin h &= \sin A \; \sin b \\
&= \sin 23.0^{\circ} \sin 30^{\circ} \\
&= 0.19556
\end{align}$$ so $h = 0.1969 \text{ radians} = 11.3^{\circ}$, and the distance from $P$ to $V$ is $0.1969 \times 6370 = 1254$ km.
To find the longitude of $V$, we first find the angle $APB$.
$$\begin{align}
\cos APB &= \frac{\tan h}{\tan b} \\
&= \frac{\tan 11.3^{\circ}}{\tan 42^{\circ}} \\
&= 0.2215
\end{align}$$ so angle $APB = 77.2^{\circ}$, and the latitude of point V is $77.2^{\circ} - 122^{\circ} = -44.8^{\circ} = 44.8^{\circ} W$
