Question

In the diagram, point O is the center of the circle and AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E. If m∠AOB = 90° and

Answer 1

Answer:

53

Step-by-step explanation:

I put in 53 on PLATO and it looks like it was the answer.

Answer 2

Answer: $53^{\circ}$

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°