digression - circumradius and the sine rule

 In euclidean geometry, the radius r of the circumcircle is given by 2r = a/sin(A), or either of the other ratios in the Sine Rule. In hyperbolic geometry, we saw that sin(A) = Δ/sinh(b)sinh(c), so that, (*) Δ = sinh(b)sinh(c)sin(A). Once again the circumradius is related to the ratios which occur in the Sine Rule. tanh(r) = 4sinh(½a)sinh(½b)sinh(½c)/Δ(cosh(a),cosh(b),cosh(c)) If we now use the result (*) for Δ, we get, using sinh(x)=2sinh(½x)cosh(½x) tanh(r) = (sinh(½a)/(cosh(½b)cosh(½c)sin(A)) This gives a neat formula for tanh(r) with no roots required. Also, we know that sinh(a)/sin(A) = sinh(b)/sin(B) = sinh(c)/sin(C) = sinh(a)sinh(b)sinh(c)/Δ Then we have, provided that the circle has a hyperbolic circumcircle of hyperbolic radius r, sinh(a)/sin(A) = sinh(b)/sin(B) = sinh(c)/sin(C) = 2tanh(r)/(cosh(½a)cosh(½b)cosh(½c). Note that, for any hyperbolic triangle, the hyperbolic area D satisfies tan(D/2) = Δ/(1+cosh(a)+cosh(b)+cosh(c)), so that sinh(a)/sin(A) = sinh(b)/sin(B) = sinh(c)/sin(C) = sinh(a)sinh(b)sinh(c)/tan(D/2)(1+cosh(a)+cosh(b)+cosh(c)).