The value of <em>x</em> where by exterior angles theorem, ∠DCO equals 2·x, 69° equals the sum of ∠DCO and x° is <em>x </em>= 23°
<h3>What is the exterior angles theorem?</h3>
The exterior angles theorem states that the exterior angle of a triangle is equal to the sum of the opposite interior angles.
The given parameters are;
Points on the circle are; ABCD
Straight lines = AOBE and DCE
Segment CO = Segment CE
∠AOD = 69°
∠CEO = x°
∠OCD = ∠ODC base angles of isosceles triangle
69° = ∠x° + ∠ODC exterior angle of a triangle
∠DOC = 180 - 2×∠ODC (Angle sum property of a triangle)
180 - 2×∠ODC + x° + 69° = 180° (Sum of angles on a straight line are supplementary)
x° + 69° - 2×∠ODC = 0
∠ODC = 2·x° (exterior angle of a triangle is equal to the sum of the opposite interior angles)
69° = x° + ∠ODC = x° + 2·x° = 3·x° (substitution property of equality)
69° = 3·x°
x = 69° ÷ 3 = 23°
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