The inradius R of the hyperbolic ΔABC satisfies
(1) tanh(R) = sinh(s-a)tan(½α) = sinh(s-b)tan(½β) = sinh(s-c)tan(½γ), and
(2) tanh(R) = Δ/2sinh(s),
(3) tanh(R) = Φ/4cos(½α)cos(½β)cos(½γ).
proof
If we apply each part of (1) and multiply by sinh3(s), we get
(3) follows from (2), the s-theorem and (R1).
We use the diagram on the right, even though it (appears) euclidean.
ΔA'BX has a right angle at A', <A'BX = ½β, d(X,A') = R, d(B,A') = s-b,
the last by the lemma. Applying the Tangent Formula,
tan(½β) = tanh(R)/sinh(s-b). This gives one part of (1).
The others are similar.
sinh3(s)tanh3(R) = sinh(s-a)tan(½α)sinh(s-b)tan(½β)sinh(s-c)tan(½γ)sinh3(s).
If we apply (Δ1),(R1), the s-theorem and standard trigonometry, we get
sinh3(s)tanh3(R) = Δ3/8. (2) follows at once.
