In general hyperbolic triangles, we found some special results for
right-angled hyperbolic triangles. These are given on the right for a
hyperbolic triangle ΔABC with a right angle at A, sides of hyperbolic
length a, b and c and angles B at B, C at C.
In euclidean geometry, if ΔABC has a right angle at A, then the angles
at B and C satisfy B+C = π/2. It follows at once that cos(C)/sin(B) = 1,
cos(B)/sin(C) = 1, tan(C)tan(B) = 1 and tan((π/2-(B+C))/2) = 1.
The angle π/2-(B+C) is actually quite useful in work on hyperbolic areas!
There are analogues in hyperbolic geometry, though since we now have
merely that B+C < π/2, we cannot expect constants on the right.
Theorem
If the hyperbolic triangle ΔABC has a right angle at A, then
(1) cos(C)/sin(B) = cosh(c),
(2) cos(B)/sin(C) = cosh(b),
(3) tan(B)tan(C) = 1/tanh(a),
(4) tan((π/2-(B+C))/2) = tanh(b/2)tanh(c/2).
proof
We use freely the fact that tanh(x) = sinh(x)/cosh(x).
(1) From the sine and cosine formulae cos(C)/sin(B) = cosh(a)/cosh(b).
From Pythagoras's Theorem, cosh(a)/cosh(b) = cosh(c), as required.
(2) and (3) are similar.
(4) This needs more work. First, we note that
(i) tan2(d/2) = (1-cos(d))/(1+cos(d)), and
(ii) tanh2(x/2) = (cosh(x)-1)/(cosh(x)+1).
From (1),(2) and (ii),
tanh2(b/2)tanh2(c/2)
= (cos(C)-sin(B))(cos(B)-sin(C))/(cos(C)+sin(B))(cos(B)+sin(C))
= (1-sin(B+C))/(1+sin(B+C)) (by an obscure identity)
= (1-cos(D))/(1+cos(D))
(where D = π/2-(B+C))
= tan2(D/2).
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Pythagoras's Theorem
The h-triangle ABC has a right angle at A
if and only if cosh(a) = cosh(b)cosh(c).
If ΔABC has a right angle at A, we have
the cosine formula
cos(B) = tanh(c)/tanh(a), and
cos(C) = tanh(b)/tanh(a).
the sine formula
sin(B) = sinh(b)/sinh(a), and
sin(C) = sinh(c)/sinh(a).
the tangent formula
tan(B) = tanh(b)/sinh(c), and
tan(C) = tanh(c)/sinh(b).
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