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A nondegenerate convex regular spherical $n$-gon contained in an open hemisphere has equal interior angles $\alpha$ in the displayed interval. The spherical excess formula gives area $n\alpha -(n-2)\pi >0$. Conversely, put equally spaced vertices at colatitude $t\in (0,\pi /2)$ around a pole. Dividing each central triangle in half gives $\cot (\alpha /2)=\cos t\tan (\pi /n)$ by the spherical law of cosines. As $t$ increases, the angle ranges continuously from the Euclidean limiting angle to $\pi$.

Work first on the unit sphere. A spherical lune of angle $\alpha$ has area $2\alpha$: its fraction of sphere area is $\alpha /(2\pi )$. The three great circles extending the sides of a nondegenerate spherical triangle divide the sphere into four antipodal pairs of triangular regions. Call the desired region $T$, of area $A$, and its three adjacent regions $T_1,T_2,T_3$, of areas $A_1,A_2,A_3$. These four regions contain one region from each antipodal pair, so $A+A_1+A_2+A_3=2\pi$. The lune at each vertex consists of $T$ and one of these adjacent regions. Therefore, if the three interior angles are $\alpha ,\beta ,\gamma$,which proves the spherical excess formula, the spherical-triangle form of the Gauss-Bonnet theorem:For a sphere of radius $s$, multiply this area by $s^2$.

Triangulate a convex spherical $n$-gon by diagonals from a vertex. There are $n-2$ spherical triangles, and their angles add to the polygon's interior angle sum. Thus its unit-sphere area isFor a nondegenerate convex regular polygon in an open hemisphere, area is positive and each angle is less than $\pi$, giving $(n-2)\pi /n<\alpha <\pi$. Every angle in this interval occurs. Indeed place $n$ vertices at equal longitude spacings and colatitude $t\in (0,\pi /2)$ about a pole, and join consecutive vertices by shorter great circle arcs. The polygon is convex and regular. Split a central isosceles spherical triangle along the perpendicular to its edge. The resulting right spherical triangle has angles $\pi /n$, $\alpha /2$, $\pi /2$, and hypotenuse $t$. The angular spherical law of cosines yieldsso $\cot (\alpha /2)=\cos t\tan (\pi /n)$. As $t$ increases from $0$ to $\pi /2$, $\alpha$ increases continuously through exactlyThe endpoints are degenerate limits, not strict convex polygons.

For the spherical triangle reflection group, positive area requires $1/p+1/q+1/r>1$. Each angle is less than $\pi$, so the integer denominators are at least two. Arrange $p\le q\le r$. If $p\ge 3$ the sum is at most one, hence $p=2$. If $q\ge 4$ the sum is again at most one; thus either $q=2$ and any $r\ge 2$ works, or $q=3$ and $r=3,4,5$. These triangles exist, for example as the reflection chambers of a regular dihedral configuration, tetrahedron, cube, and dodecahedron, respectively.

To count group elements using the assumed tessellation, first label the original triangle's three vertices by their types and propagate these labels through side reflections. Around a type-$p$ vertex there are $2p$ sectors; the associated alternating reflections make a full turn and return the labels unchanged. The same holds at the other types. Every closed path in the dual tessellation is generated by these circuits around vertices, since the sphere is simply connected. Thus the labels are globally consistent, including when two angle denominators coincide. Every generated isometry preserves these labels. An element stabilizing a triangle therefore fixes its three vertices and is the identity. Consequently group elements correspond bijectively to triangles, and area givesThe possibilities, including permutations of the entries, are

Here is an explicit dodecahedron construction for the last case, avoiding an assumption about which vertices of the tessellation to join. Put $\phi =(1+\sqrt{5})/2$, and take the convex hull of the twenty pointswith independent signs. They all have squared radius three. The supporting plane $\phi y+z={\phi }^2$ contains the five verticesin cyclic order. Consecutive distances are $2/\phi$, and nonconsecutive distances are $2$, so this is a regular pentagon. Changing the two signs in its normal $(0,\phi ,1)$ and cyclically permuting coordinates gives twelve supporting planes, all with a congruent pentagon. Each listed vertex lies on three of these faces; the sixty face-edge incidences pair into thirty edges. The faces form a closed convex polyhedral surface with $20-30+12=2$, hence exhaust the boundary of the convex hull. This constructs a regular dodecahedron.

Choose the spherical directions of a face center $F=(0,\phi ,1)$, the midpoint $M=(0,\phi ,0)$ of its edge between $(\pm {\phi }^{-1},\phi ,0)$, and that edge's endpoint $V=({\phi }^{-1},\phi ,0)$. Normalize these to the unit sphere. The three side planes of triangle $FMV$ are $x=0$, $z=0$, and $\mathbf{n}\cdot \mathbf{x}=0$ with $\mathbf{n}=(-\phi ,{\phi }^{-1},-1)$. Their reflections are coordinate sign changes in the first two cases andin the third, since $\Vert \mathbf{n}{\Vert }^2=4$. Substitution in the twenty listed vertices, using ${\phi }^2=\phi +1$, shows that all three reflections permute the vertex set. The side-plane angles give interior angles $\pi /2$ at $M$, $\pi /3$ at $V$, and $\pi /5$ at $F$ because $|n_z|/\Vert \mathbf{n}\Vert =1/2$ and $|n_x|/\Vert \mathbf{n}\Vert =\phi /2=\cos (\pi /5)$. Thus this triangle is congruent to $\Delta$, and its side-generated group is $G$, conjugated by the aligning isometry.

It already has $120$ elements. Conversely, a symmetry of the dodecahedron is determined by the image of one face, one vertex on that face, and the orientation along its boundary: there are at most $12\cdot 5\cdot 2=120$ choices. An isometry fixing these choices fixes that face pointwise and must preserve the polyhedron's interior side, hence is the identity. Therefore the complete symmetry group of the constructed dodecahedron is $G$, including orientation-reversing symmetries.

Use suffix notation, summing repeated indices from one to three. The contraction identity for the Levi-Civita symbol givesHere $a\cdot a=1$ because $a$ is a unit vector. For the second cross product identity, use ${ϵ}_{ijk}{ϵ}_{j\ell m}={\delta }_{im}{\delta }_{k\ell }-{\delta }_{i\ell }{\delta }_{km}$:The first term vanishes because $a_{\ell }{a}_n$ is symmetric while the Levi-Civita symbol is antisymmetric in those indices. The last equality uses cyclicity and antisymmetry of the scalar triple product.

On the unit sphere, a great circle has radius one, so the spherical distance is its shorter arc length, equal to its central angle in $[0,\pi ]$. The dot product of its endpoint unit vectors is the cosine of that angle. HenceFor the angular formulas, take the ordinary nondegenerate convex minor spherical triangle: no pair of vertices is antipodal and the three vertices do not lie on one great circle. These are needed for the displayed denominators and interior angles to be defined. Merely requiring three distinct points, as the PDF does, does not exclude degeneracy.

Put $p=\delta (A,B)$, $q=\delta (A,C)$ and $r=\delta (B,C)$. The outgoing unit tangent vectors at $A$ areTheir dot product is $\cos \alpha$. Equivalently, they are obtained from the normalized plane normals $a\times b$ and $a\times c$ by the same right-angle rotation in the tangent plane: $u=(a\times b)\times a/|a\times b|$, and similarly for $v$. Rotation preserves the dot product, givingSince $|a\times b|=\sin p$ and $|a\times c|=\sin q$, the first proved identity becomes the spherical law of cosines:For the spherical law of sines, take norms in the second proved identity. Writing $\tau =a\cdot (b\times c)$, and using $0<\alpha <\pi$, givesCyclic permutation gives the corresponding formulas at $B$ and $C$. Dividing each by the positive product $\sin p\sin q\sin r$ provesFinally, if all sides equal $s$, the spherical law of cosines givesbecause $0<s<\pi$ makes $1+\cos s>0$ and $\cos s<1$. Cosine is strictly decreasing on $(0,\pi )$, so every angle of a nondegenerate equilateral spherical triangle is greater than $\pi /3$.

For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.

A spherical triangle is bounded by three great circle arcs. For the usual convex minor spherical triangle, the vertices are neither antipodal nor on one great circle, and its angular sides and interior angles lie in $(0,\pi )$. An interior angle is measured between outgoing tangent vectors to the two sides at a vertex, rather than between straight chords inside the sphere. The spherical law of sines and spherical law of cosines determine its side-angle relations; the spherical excess formula determines its area.

Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.