In a right-angled triangle ABC, the center O of the inscribed circle is connected to the vertex of the right angle C

In a right-angled triangle ABC, the center O of the inscribed circle is connected to the vertex of the right angle C and to the vertex B. It is known that AB = 36 is the angle BOC 105 degrees. Find the radius of the inscribed circle.

From the center O of the circle, draw the perpendiculars OH and OK to the legs of the triangle.

By the property of tangents to the circle, CH = CK, and OH = OK as the radii of the circle, since the angle is C = 900, and OH and OK are perpendiculars, then OНСН is a square. Then CO is the diagonal of the square, and the angle ОСК = 45. Then the angle OBC = 180 – 45 – 105 = 30.

Since point O is the intersection point of the bisectors of the triangle ABC, then the angle ABC = 2 * OBC = 2 * 30 = 60, then the angle BAC = 180 – 90 – 60 = 30.

The BC leg lies opposite the angle 30, then its length is equal to half the length of AB.

BC = 36/2 = 18 cm.

By the Pythagorean theorem, we determine the length of the leg AC.

AC^2 = AB^2 – CB^2 = 1296 – 324 = 972.

AC = 18 * √3 cm.

We define the radius of the inscribed circle through the lengths of the sides of the triangle.

R = (AC + BC – AB) / 2 = (18 * √3 + 18 – 36) / 2 = 9 * √3 + 9 – 18 = 9 * (√3 + 1 – 2) = 9 * (√3 – 1) see.

Answer: The radius of the circle is 9 * (√3 – 1) cm.