The tangents at two points A and B on the circle with centre O intersect at P. If in quadrilateral PAOB, `angle AOB : angle APB = 5 : 1`, then measure of `angle APB` is :

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `15^(@)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a circle with center O, and two tangents PA and PB drawn from an external point P. The angles we need to consider are ∠AOB and ∠APB, and we know the ratio of these angles is 5:1.

2. **Setting Up the Angles**:
Let ∠APB = k. According to the given ratio, ∠AOB = 5k.

3. **Using the Properties of Tangents**:
Since PA and PB are tangents to the circle at points A and B respectively, we know that:
- ∠PAO = 90° (the angle between the radius OA and the tangent PA)
- ∠PBO = 90° (the angle between the radius OB and the tangent PB)

4. **Sum of Angles in Quadrilateral PAOB**:
The sum of the angles in a quadrilateral is 360°. Therefore, we can write the equation:
\[ \angle PAO + \angle PBO + \angle APB + \angle AOB = 360° \]
Substituting the known values, we have:
\[ 90° + 90° + k + 5k = 360° \]

5. **Simplifying the Equation**:
Combine like terms:
\[ 180° + 6k = 360° \]

6. **Isolating k**:
Subtract 180° from both sides:
\[ 6k = 360° - 180° \]
\[ 6k = 180° \]

7. **Solving for k**:
Divide both sides by 6:
\[ k = \frac{180°}{6} = 30° \]

8. **Finding ∠APB**:
Since ∠APB = k, we find:
\[ \angle APB = 30° \]

### Conclusion:
The measure of ∠APB is **30 degrees**.