PA and PB are tangents to a circle with centre 0, from a point P outside the circle, and A and B are point on the circle. If `angleAPB = 30^(@)`, then `angleOAB` is equal to:

Options

A. `25^(@)`
B. `40^(@)`
C. `15^(@)`
D. `50^(@)`

Answer

To find the angle \( \angle OAB \) given that \( \angle APB = 30^\circ \), we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center \( O \) and point \( P \) outside the circle from which two tangents \( PA \) and \( PB \) touch the circle at points \( A \) and \( B \) respectively. The angle \( \angle APB \) is given as \( 30^\circ \).

### Step 2: Identify Right Angles
Since \( PA \) and \( PB \) are tangents to the circle, the radius \( OA \) is perpendicular to the tangent \( PA \) at point \( A \), and the radius \( OB \) is perpendicular to the tangent \( PB \) at point \( B \). Therefore, we have:
\[ \angle OAP = 90^\circ \quad \text{and} \quad \angle OBP = 90^\circ \]

### Step 3: Analyze Quadrilateral \( OAPB \)
In quadrilateral \( OAPB \), the sum of the angles is \( 360^\circ \). We can express this as:
\[ \angle OAP + \angle APB + \angle OBP + \angle OAB = 360^\circ \]
Substituting the known values:
\[ 90^\circ + 30^\circ + 90^\circ + \angle OAB = 360^\circ \]
This simplifies to:
\[ 210^\circ + \angle OAB = 360^\circ \]

### Step 4: Solve for \( \angle OAB \)
Now, we can solve for \( \angle OAB \):
\[ \angle OAB = 360^\circ - 210^\circ = 150^\circ \]

### Step 5: Analyze Triangle \( AOB \)
In triangle \( AOB \), we know that:
\[ \angle AOB + \angle OAB + \angle OBA = 180^\circ \]
Let \( \angle OAB = x \) and \( \angle OBA = x \) (since they are equal). Then we have:
\[ 150^\circ + x + x = 180^\circ \]
This simplifies to:
\[ 150^\circ + 2x = 180^\circ \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ 2x = 180^\circ - 150^\circ = 30^\circ \]
\[ x = \frac{30^\circ}{2} = 15^\circ \]

### Conclusion
Thus, the angle \( \angle OAB \) is:
\[ \angle OAB = 15^\circ \]