Answer: [tex]53^{\circ}[/tex]
Step-by-step explanation:
Given : Point O is the center of the circle, AC and BD are the chord of the circle, E is the intersection point of AC and BD,
m∠AOB = 90° and m∠COD = 16°
We have to find : m∠CED
Since, By joining point B and C (construction)
m∠AOB = 90° ⇒ m∠ACB = 45° (by the center angle theorem)
Similarly, by joining points A and D,
m∠AOB = 90° ⇒ m∠ADB = 45°
Since, triangles COD and CBD are made by the same arc CD inside the circle having the center O.
Thus, m∠CBD = m∠COD/2 = 16/2 = 8°
⇒ m∠CBD = 8°
But, m∠CED = m∠CBD + m∠ACB (by the property of exterior angle of triangle)
⇒ m∠CED = 8° + 45° = 53°
Therefore, m∠CED = 53°
