The acute angle of a right-angled triangle is 30 °. At what angle is each leg visible from the center of the circle

The acute angle of a right-angled triangle is 30 °. At what angle is each leg visible from the center of the circle described around this triangle?

Let △ ABC be inscribed in a circle centered at point O, ∠C = 90 °, ∠A = 30 °, AB – hypotenuse, AC and BC – legs.
1. By the theorem on the sum of the angles of a triangle:
∠А + ∠В + ∠С = 180 °;
30 ° + ∠В + 90 ° = 180 °;
∠В = 180 ° – 120 °;
∠В = 60 °.
2. Since ∠С is equal to 90 °, then it rests on an arc equal to 180 ° (the inscribed angle is equal to half of the arc on which it rests). An arc equal to 180 ° is pulled together by the diameter, thus the diameter of the circle is also the hypotenuse AB △ ABC. Point O divides AB in half into two radii: OA = OB.
From point O we draw a segment to point C: OS is the radius, therefore OA = OB = OS.
3. Consider △ AOC: OA = OC ⇒ △ AOC is isosceles, and ∠CAO and ∠ACO are the angles at the base of the AC.
∠САО (aka ∠А) = ∠АСО = 30 °.
By the theorem on the sum of the angles of a triangle:
∠САО + ∠АОС + ∠АСО = 180 °;
30 ° + ∠АС + 30 ° = 180 °;
∠АС = 180 ° – 60 °;
∠АС = 120 °.
Thus, the leg AC from the center of the circle O is visible at an angle of 120 °.
4. Consider △ OWS: OS = OV ⇒ △ OWS is isosceles, and ∠ОСВ and ∠ОВС are angles at the base of the BC.
∠ОСВ = ∠ОВС (aka ∠В) = 60 °.
By the theorem on the sum of the angles of a triangle:
∠ОСВ + ∠СОВ + ∠ОВС = 180 °;
60 ° + ∠СОВ + 60 ° = 180 °;
∠СОВ = 180 ° – 120 °;
∠СОВ = 60 °.
Thus, the leg BC from the center of the circle O is seen at an angle of 60 °.
Answer: ∠AOC = 120 °, ∠СОВ = 60 °.