= Spherical geometry
{wiki}
<Spherical geometry> studies <lengths> and <angles> measured along the surface of a <sphere>, rather than through its interior. It is a useful model whenever travel is constrained to a round surface. Its straightest paths are <great circle> arcs: they are <geodesics>, even though they appear curved in the surrounding <Euclidean space>. Two distinct <great circles> meet at an <antipodal pair>, so the parallel-line picture of <Euclidean geometry> does not apply.
For a <sphere> of radius $R>0$, write two positions as $Ru,Rv$, with $u,v$ <unit vectors>. Their shorter surface route has <great-circle distance>
$$
d=R\arccos(u\cdot v),
$$
where the <dot product> gives the central <angle> in <radians>. The straight chord through the surrounding <Euclidean space> instead has <length> $2R\sin(d/(2R))$. For a quarter-circle separation these are $\pi R/2\approx1.571R$ and $\sqrt2R\approx1.414R$: specifying which distance we mean has a measurable effect. On the <unit sphere>, $R=1$ and <great-circle distance> equals the central <angle>.
Consider the <spherical triangle> with vertices $(R,0,0)$, $(0,R,0)$ and $(0,0,R)$, joined by the shorter <great circle> arcs. Each side has <length> $\pi R/2$. At each vertex the two outgoing <tangent vectors> point along the other coordinate axes and are <orthogonal>, so all three interior <angles> are $\pi/2$. This is the positive coordinate octant of the <sphere>, not a flat <triangle> drawn inside it.
For a convex minor <spherical triangle> with interior <angles> $\alpha,\beta,\gamma$ in <radians>, the <spherical excess formula> states
$$
\mathcal A=R^2(\alpha+\beta+\gamma-\pi).
$$
Our octant therefore has <area> $R^2(3\pi/2-\pi)=\pi R^2/2$. Independently, eight congruent octants partition the <sphere>, giving $4\pi R^2/8$ again. Thus <angles> alone determine <area> when $R$ is fixed, unlike <Euclidean geometry>, where <similar triangles> can have arbitrarily different <areas>. The <Gauss-Bonnet theorem> explains the connection: the <sphere>'s constant <Gaussian curvature> $1/R^2$ times the <area> equals this excess. For <spherical triangles> whose <area> is small compared with $R^2$, the excess is small, recovering the familiar flat picture.
\Image[/spherical-triangle-octant.png]
{title=Three quarter-great-circle sides enclose a spherical octant with three right angles and area $\pi R^2/2$}
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