Hyperbolic triangle identities and angle defect

Try

In a hyperbolic right triangle with leg lengths a and b satisfying cosh a=3/2 and cosh b=2, the hypotenuse satisfies cosh c=3. The relation is multiplicative in hyperbolic cosines, not c²=a²+b² exactly.

All lengths here use the curvature −1 normalization. Changing the curvature scale changes the numerical length units, so formulas must carry that convention. The hyperbolic functions are defined by cosh t=(eᵗ+e⁻ᵗ)/2 and sinh t=(eᵗ−e⁻ᵗ)/2.

Predict

Predict whether small hyperbolic triangles resemble Euclidean ones. Since cosh t=1+t²/2+O(t⁴), the right-triangle identity reduces to Euclidean Pythagoras at leading order. Approximate local agreement does not make the exact global geometries identical.

Predict whether two hyperbolic triangles with the same three angles can have arbitrarily different sizes, as Euclidean similar triangles can. Hyperbolic angle identities determine the side lengths once curvature is fixed.

Experiment

Compare exact and approximate relations.

Geometry	Right-triangle relation	Normalization
Euclidean	c²=a²+b²	Ordinary lengths
Hyperbolic	cosh c=cosh a cosh b	Curvature −1
Small hyperbolic triangle	c²≈a²+b²	Higher-order error omitted

Use a=b and expand both sides through quadratic order. Explain why the approximation is legitimate only as the lengths tend to zero, rather than assigning it an unspecified numerical accuracy at every scale.

Understand

The Lorentz model derives the cosine law. Place P=(1,0,0), Q=(cosh c,sinh c,0), and R=(cosh b,sinh b cos A,sinh b sin A). These lie on the upper hyperboloid. The tangent directions at P make angle A in its positive tangent metric. The side opposite A has length a=d(Q,R).

Using cosh a=−⟨Q,R⟩L gives cosh a=cosh b cosh c−sinh b sinh c cos A. When A=π/2, the last term vanishes, yielding the right-triangle relation.

Solving the cyclic side identities yields the angle cosine rule cosh a=(cos A+cos B cos C)/(sin B sin C). This algebraic consequence uses nondegenerate interior angles in (0,π). Its denominator is positive, and cosh is strictly increasing on positive lengths. Thus equal corresponding angle triples force equal side triples; there is no freely rescalable family at fixed curvature.

Apply

For a simple geodesic triangle bounding a disk, Gauss–Bonnet gives ∫K dA plus the sum of exterior turning angles equal to 2π. With K=−1 and exterior turns π−A, π−B, π−C, this becomes Area=π−A−B−C. The positivity of area implies angle sum below π. This lesson applies Gauss–Bonnet as an explicitly named theorem; its general proof belongs to differential geometry.

Repair “The defect is the area on every hyperbolic plane regardless of units.” If curvature is −κ², area is defect divided by κ². Angle defect has no length unit, while area does.

For optional depth, the angle formula’s derivation can be checked by substituting the side cosine law and simplifying with cosh²−sinh²=1. The model computation and the area theorem provide different kinds of evidence and should not be conflated.

Continue

Keep curvature normalization, nondegeneracy and theorem dependencies beside every triangle calculation. Continue to spherical geometry to compare the sign changes that arise with positive curvature.