Points A,B,C and D are concyclic. BC is the diameter and O is the centre of the circle, `angleBAO=69^@` find `angleDAO+angleDCO`

Options

A. A)138
B. B)111
C. C)121
D. D)none of these

Answer

To solve the problem, we need to find the sum of angles \( \angle DAO + \angle DCO \) given that points A, B, C, and D are concyclic, with BC as the diameter and O as the center of the circle. We know that \( \angle BAO = 69^\circ \).

### Step-by-step Solution:

1. **Identify the Circle and Points**:
- Draw a circle with points A, B, C, and D on its circumference.
- Mark BC as the diameter of the circle, and O as the center of the circle.

2. **Label the Angles**:
- We know that \( \angle BAO = 69^\circ \).

3. **Determine \( \angle ABO \)**:
- Since OA and OB are radii of the circle, triangle AOB is isosceles.
- Therefore, \( \angle ABO = \angle OAB = 69^\circ \).

4. **Calculate \( \angle AOB \)**:
- The sum of angles in triangle AOB is \( 180^\circ \).
- Thus, \( \angle AOB = 180^\circ - \angle BAO - \angle ABO = 180^\circ - 69^\circ - 69^\circ = 42^\circ \).

5. **Find \( \angle DAB \)**:
- Since A, B, C, D are concyclic and BC is the diameter, \( \angle BAC = 90^\circ \) (angle subtended by diameter).
- Therefore, \( \angle DAB + \angle BAC = 180^\circ \) (opposite angles in cyclic quadrilateral).
- Hence, \( \angle DAB = 180^\circ - 90^\circ = 90^\circ \).

6. **Calculate \( \angle DCO \)**:
- Since \( \angle DAB = 90^\circ \) and \( \angle BAO = 69^\circ \), we can find \( \angle DAB \) using the cyclic property.
- \( \angle DCO = \angle DAB - \angle BAO = 90^\circ - 69^\circ = 21^\circ \).

7. **Find \( \angle DAO + \angle DCO \)**:
- We know \( \angle DAO = \angle DAB = 90^\circ \).
- Therefore, \( \angle DAO + \angle DCO = 90^\circ + 21^\circ = 111^\circ \).

### Final Answer:
Thus, \( \angle DAO + \angle DCO = 111^\circ \).

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