In a right-angled triangle ABC (C = 90 °), the bisectors CD and AE meet at point

In a right-angled triangle ABC (C = 90 °), the bisectors CD and AE meet at point O. The angle AOC is 115 °. Find its smaller acute angle of triangle ABC.

Consider a triangle ABC, <C = 90º. СD – bisector <C, AE – bisector <А.
By the property of bisectors:
<C / 2 = <АCD, <ВCD = 90/2 = 45º.
<A / 2 = <EAC, <BAE.
<АCD = <АCO.
<EAC = <OAC.
Consider a triangle АСО, <СОА = 115º, <АСО = 45º, find the angle <ОАC.
By the property of the angles of a triangle:
<СОА + <АСО + <ОАC = 180º
<ОАC = 180- <СОА + <АСО = 180º-115º-45º = 20º.
Let’s return to the triangle ABC, define <A:
<ОАC = <ЕАC = <А / 2
Where:
<А = 2 * <ОАC = 2 * 20 = 40º.
By the property of the angles of a triangle:
<A + <B + <C = 180º.
<B = 180- <A- <C = 180º-40º-90º = 50º.
Answer: the smaller angle of the triangle ABC is <A = 40º.