A chord of a circle is equal to the radius of the circle. Find the angle substended by the chords at a point on the minor arc and also at a point on the major arc.
We have,
Radius OA = Chord AB
OA = OB = AB
Then, triangle OAB is an equilateral triangle
Therefore,
∠AOB = 60o (Angle of an equilateral triangle)
By degree measure theorem,
∠AOB = 2 ∠APB
60o = 2 ∠APB
∠APB = 30o
Now,
∠APB + ∠AQB = 180o (Opposite angle of cyclic quadrilateral)
30o + ∠AQB = 180o
∠AQB = 150o
Therefore,
Angle by chord AB at minor arc = 150o
And, by major arc = 30o