Two chords AB and CD of a circle with centre o intersect each other at the point P. If `/_AOD = 20^@` and `/_BOC = 30^@`, then `/_BPC `is equal to:

Options

A. `50^@`
B. `20^@`
C. `25^@`
D. `30^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the angles given in the problem.
We are given:
- Angle AOD = 20°
- Angle BOC = 30°

### Step 2: Use the property of angles subtended by arcs at the center and at the circumference.
According to the property, the angle subtended by an arc at the center is double the angle subtended at any point on the circle.

- For arc AD:
- Angle AOD = 20° (at the center)
- Therefore, angle ACP (subtended at point P) = 20° / 2 = 10°

- For arc BC:
- Angle BOC = 30° (at the center)
- Therefore, angle BAP (subtended at point P) = 30° / 2 = 15°

### Step 3: Apply the exterior angle theorem in triangle BPC.
In triangle BPC, we can use the exterior angle theorem which states that the exterior angle (angle BPC) is equal to the sum of the two opposite interior angles (angles ACP and BAP).

- Angle BPC = Angle ACP + Angle BAP
- Angle BPC = 10° + 15° = 25°

### Conclusion:
Thus, the angle BPC is equal to 25°.