Find the length of a circle inscribed in a right-angled triangle with a hypotenuse C and an acute angle a.

Let’s designate a triangle as ABD, angle A is a straight line, angle D is equal to a, then angle B is equal to (90 – a).
Then, by the sine theorem:
X = AB = C * sina.
Y = AD = C * sin (90-a) = C * cosa.

The circumference is equal to:
P = 2 * π * r.
Find the radius of the circle.

Area of ​​a triangle in terms of radius:
S = r * p, r – radius, p – semi-perimeter.
r = S / p.
On the other side
S = (p * (p-X) * (p-Y) * (p-C)) ^ (1/2),
p = (X + Y + C) / 2 = (C * sina + C * cosa + C) / 2 = C / 2 * (sina + cosa + 1).
p – C = C / 2 * (cosa + sina – 1);
p – X = C / 2 * (1 + cosa – sina);
p – Y = C / 2 * (1 + sina – cosa) = C / 2 * (1 – (cosa – sina));
p * (pX) * (pY) * (pC) = C ^ 4/16 * (sina + cosa + 1) * (cosa + sina – 1) * (1 + cosa – sina) * 1 – (cosa – sina ) = (C ^ 4/16) * ((sina + cosa) ^ 2 – 1) * (1 – (cosa – sina) ^ 2) = (C ^ 4/16) * (sin ^ 2 a + cos ^ 2 a + 2cosa * sina – 1) * (1 – cos ^ 2 a + 2cosa * sina – sin ^ 2 a) = (C ^ 4/16 * 2cosa * sina * 2cosa * sina = (C ^ 4/16) * (2cosa * sina) ^ 2;
S = (C ^ 4/16 * (2cosa * sina) ^ 2) ^ (1/2) = (C ^ 2/4) * 2cosa * sina = (C ^ 2 * cosa * sina) / 2,
r = S / p = ((C ^ 2 * cosa * sina) / 2) / C / 2 * (sina + cosa + 1) = C * cosa * sina / (sina + cosa + 1).
Thus, the circumference is:
P = 2 * π * r = 2 * π * C * cosa * sina / (sina + cosa + 1) = π * C * sin2a / (sina + cosa + 1).