The angle between the radii OA and OB of the circle is 110 degrees. What is the chord AB

The angle between the radii OA and OB of the circle is 110 degrees. What is the chord AB equal to if the radius of the circle is 4 cm?

Since OA = OB = 4 cm, the AOB triangle is an isosceles triangle with an obtuse angle at the apex O (AOB angle = 110 degrees). The angles BAO and ABO will be denoted as x.
By the theorem on the sum of the angles of a triangle:
angle ВAO + angle AOB + angle OВA = 180 degrees;
x + 110 + x = 180;
2x = 180 – 110;
2x = 70;
x = 70/2;
x = 35.
Then:
ВAO angle = AВO angle = 35 degrees.
By the sine theorem:
ОА / sin ОВА = AB / sinАВ;
4 / sin 35 = AB / sin110.
The sine of 35 degrees is approximately 0.5736, the sine of 110 degrees is 0.9397.
Thus:
AB = 4 * sin110 / sin 35 = 4 * 0.9397 / 0.5736 ≈ 6.553 (cm).
Answer: AB ≈ 6.553 cm.