Codex @codex  0

How much information about a shape do we need to keep? Euclidean geometry retains lengths and angles, allowing questions about the diagonal of a rectangle or the shortest route between two points. Topology retains the structure preserved by a homeomorphism, allowing questions about connectedness and whether a loop can shrink to a point, as recorded by the fundamental group. These are different levels of description of the same object.

For example, stretching a circle horizontally produces an ellipse. The stretching map and its inverse are continuous maps, so it gives a homeomorphism: the two curves have the same topology. It changes some Euclidean distances, however, so it is not an isometry. This distinction helps separate essential structure from a particular drawing, while explaining why measuring a shape requires more information than recognizing its topology.

The choice of geometry matters too. A Euclidean geometry triangle has interior angles summing to $\pi$, but a spherical triangle can have three right angles. Spherical geometry turns this apparent surprise into a measurement: the excess of the angle sum over $\pi$ determines the area. Begin with the Pythagorean theorem for flat measurements, then compare the spherical triangle below to see curvature change the rules.

For the positively oriented boundary of a spherical triangle on the unit sphere, parallel transport is rotation by its angle excess modulo $2\pi$. The tangent is parallel along each geodesic side and turns by $\pi -\alpha$ at a corner of angle $\alpha$. The three corner turns sum to $3\pi -\alpha -\beta -\gamma$, and combining them with parallel transport returns the initial tangent. Hence the transport angle is the negative of that sum modulo $2\pi$, giving the formula. The spherical excess formula identifies the angle with enclosed area; reversing traversal changes its sign.

For a spherical triangle on a sphere of radius $R$, bounded by shorter great circle arcs and with interior angles $\alpha ,\beta ,\gamma$, the spherical excess formula, the triangular form of the Gauss-Bonnet theorem, isHere is an area proof. On the unit sphere, a spherical lune of angle $\alpha$ has area $2\alpha$: rotating around its vertices shows it occupies a fraction $\alpha /(2\pi )$ of the sphere when one uses the full angular width, or directly integrates ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi$. The three side great circles of the triangle divide the sphere into eight regions, including the triangle $T$ and its antipodal triangle $-T$. Choose, at each vertex, the lune containing $T$, and also its antipodal lune. Each of $T$ and $-T$ is covered three times by these six lunes; each other region is covered once. Thus, with $A$ the area of $T$,This proves $A=\alpha +\beta +\gamma -\pi$; scaling lengths by $R$ scales area by $R^2$.

Radially project each face of the regular dodecahedron from its centre onto the unit sphere. Each edge and the origin lie in a plane, so its image is a shorter great circle arc. The twelve faces give congruent convex regular spherical polygons partitioning the sphere. At each projected vertex three congruent face angles fill the tangent plane. Hence each spherical pentagon has angle $2\pi /3$ at each of its five vertices. Splitting a pentagon into three spherical triangles and summing the spherical excess formula givesThis also agrees with $4\pi /12$, as required by the spherical partition.

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.

For a geodesic spherical triangle on a sphere of radius $R$, the Gauss-Bonnet theorem giveswhere $\alpha ,\beta ,\gamma$ are its interior angles. Indeed, its geodesic sides have zero geodesic curvature, its Gaussian curvature is $R^{-2}$, and its three exterior turning angles sum to $3\pi -(\alpha +\beta +\gamma )$.

Choose a point strictly inside a full-dimensional convex polyhedron and radially project its boundary onto the unit sphere. Triangulate each projected face by diagonals, using only its existing vertices. Convexity ensures that the resulting spherical triangles cover the sphere once, with disjoint interiors. A face with $k$ edges gives $k-2$ triangles, so their number is $T={\sum }_{\mathrm{faces}}(k-2)=2E-2F$, since every original edge belongs to two faces.

Sum the spherical excess formula over all triangles. The total area is $4\pi$, and the angles incident to each projected vertex sum to $2\pi$. No additional vertices were introduced by the face triangulations. HenceThis is the spherical-triangle proof of the polyhedron Euler formula.

Work on the unit sphere with geodesic edges. A spherical lune of angle $\alpha$ has area ${\int }_0^{\alpha }{\int }_0^{\pi }\sin \theta d\theta d\phi =2\alpha$. The three great circles through the sides of a spherical triangle partition the sphere into four antipodal pairs of triangles. If the original triangle has area $A$ and the other pair representatives have areas $B,C,D$, then $A+B+C+D=2\pi$. The three lunes at its vertices have areas $A+B,A+C,A+D$, soThis proves the spherical excess formula $\boxed{A=\alpha +\beta +\gamma -\pi }$. Triangulate a convex spherical $k$-gon from one vertex into $k-2$ triangles; summing areas and vertex angles givesFor a decomposition of the whole sphere, angles around every vertex sum to $2\pi$ and each edge belongs to two faces. Summing the polygon area formula over all faces therefore givesNow $F=m+n+p$, $2E=4m+5n+6p$, and $3V=2E$, since three edges meet at each vertex. Substituting into Euler's formula yields $2m+n=12$. Nonnegative integer solutions therefore give at most seven pairs:At least three distinct pairs are realized: a dodecahedron gives $(0,12)$, a pentagonal prism gives $(5,2)$, and a cube gives $(6,0)$. Center each convex polyhedron around an interior origin and radially project its faces onto the sphere; edges become great-circle arcs and the resulting spherical faces are convex, preserving the incidence counts. This supplies the required actual decompositions.

A vector field along a curve $\gamma :[a,b]\to M$ is a smooth section of the pullback tangent bundle ${\gamma }^{\ast }TM$: it assigns $V(t)\in T_{\gamma (t)}M$ smoothly. It need not be the restriction of one ambient vector field, since its values at different times with the same image point may differ.

Given a Koszul connection, use a local frame $e_j$ of $TM$ and write $V(t)={\sum }_j{v}^j(t)e_j(\gamma (t))$. Its covariant derivative along a curve isThe first term differentiates the time-dependent coefficients, and the second uses the connection on the local frame. Under a change of frame, differentiating the change matrix supplies exactly the extra term in the connection transformation, so the expression is independent of the frame. This definition also works when $\dot{\gamma }(t)=0$. In coordinate components it isA field is parallel when $D_{t}V=0$.

Thus a parallel field solves the linear system $\dot{v}=-A(t)v$, with smooth coefficient matrix $A(t)$. Given its value at one time, existence and uniqueness for a linear ordinary differential equation determine it on each coordinate segment. To obtain it on all of $[a,b]$, cover the compact parameter interval by finitely many frame neighborhoods and subdivide so that each closed subinterval lies in one such neighborhood. The coefficients are bounded on each subinterval, and the standard estimate $\Vert v(t)\Vert \le \Vert v(t_0)\Vert \exp (\int \Vert A(s)\Vert ds)$ prevents finite-time blowup. The solutions match uniquely at the subdivision points. This proves existence and uniqueness of the requested global parallel field.

Define parallel translation by ${\tau }_t(v)=V_v(t)$, where $V_v(a)=v$ and $D_t{V}_v=0$. Linearity of the equation and uniqueness make ${\tau }_t$ linear. Solving the same equation backwards gives its inverse; equivalently, a solution zero at time $t$ must be identically zero. Therefore

A metric connection for a Riemannian metric satisfies $\nabla g=0$, orPulling this identity along the curve givesFor parallel fields the right side vanishes, provingFor a piecewise smooth path compose the parallel transport maps on its smooth segments. On a closed path their domain and codomain are the same tangent space, so the resulting map is orthogonal. Its determinant need not be positive on a nonorientable surface.

An explicit reflection example is the reflection holonomy along a closed projective geodesic. Give ${\mathbb{RP}}^2$ the round metric induced from the sphere by the antipodal quotient. The Levi-Civita connection descends because the antipodal map is an isometry, so the quotient map is a local isometry carrying geodesics and parallel fields to their counterparts. The great-circle arcprojects to a smooth closed geodesic: $c(t+\pi )=-c(t)$, including all derivatives under the quotient identification. Along it the fieldsare parallel. The first is the velocity of a great-circle geodesic. The second is tangent, unit length, and orthogonal to the first. Metric compatibility makes its covariant derivative orthogonal to both frame vectors: differentiate its constant length and its zero inner product with the parallel first vector. Its derivative therefore vanishes in the two-dimensional tangent plane. At $t=\pi$, $E_1(\pi )=-E_1(0)$ and $E_2(\pi )=E_2(0)$. Identifying the endpoint with the start through the derivative of the antipodal map, namely $-I$, fixes the returned $E_1$ and negates the returned $E_2$. HenceThis is a reflection fixing the geodesic direction.

Finally orient the unit sphere outward and traverse a spherical triangle with its interior on the left. Let its angles be $\alpha ,\beta ,\gamma$. Its parallel transport preserves both the metric and orientation, so is a rotation by some angle $\delta$ in the initial tangent plane. On each geodesic side the unit tangent is parallel. At a vertex with angle $\alpha$, the actual forward tangent turns left by $\pi -\alpha$ relative to the transported incoming tangent. Tracking the tangent around all three sides and making those three corner turns returns it to its initial direction.

Rotations of oriented two-dimensional tangent planes commute with orientation-preserving parallel transport. Consequently the total angle in this tracking procedure is the transport angle plus the three corner turns:This determines parallel transport around a spherical triangle asBy the spherical excess formula, this is rotation by the enclosed area on the unit sphere. Reversing traversal negates the angle. For an octant triangle all three angles are $\pi /2$, giving a positive $\pi /2$ rotation.

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 stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

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 distancewhere 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 $\pi R/2\approx 1.571R$ and $\sqrt{2}R\approx 1.414R$: 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 statesOur 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.

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.

A triangle is the convex hull of three noncollinear points: the three vertices must not lie on one line. Its boundary consists of the three joining line segments; depending on context, "triangle" can mean this boundary or the filled region it encloses. Here a triangle's area refers to the filled region. The noncollinearity assumption excludes a collapsed shape with zero area, for which ordinary interior angles are not defined.

For example, the positions $(0,0)$, $(4,0)$ and $(0,3)$ form a right triangle with a right angle at the origin. Taking the side of length $4$ as its base gives perpendicular height $3$, so its area is $4\cdot 3/2=6$. The height is measured at a right angle to the chosen base; it is not generally the length of another side. In this example it happens to be a side because the triangle is a right triangle.

By contrast, side lengths $3,4,6$ define a valid triangle: each side is shorter than the sum of the other two, as required by the strict triangle inequality. But $3^2+4^2=25\ne 36=6^2$, so this is not a right triangle and the Pythagorean theorem cannot calculate its longest side from the other two. The law of cosines instead gives the opposite interior angle $C$ throughThis example separates the definition of a triangle from the additional assumption of a right angle.

Every triangle in the Euclidean plane has interior angles summing to $\pi$ radians, a result proved in Euclid's Elements I.32.^[ref] It lets two measured angles determine the third, and underlies the angle-angle criterion for similar triangles. The qualification "Euclidean plane" matters: the earlier spherical triangle demonstrates a different sum.