Two circles with centres P and Q intersect at B and C, A, D are points on the circles with centres P and Q respectively such that A, C, D are colliner. If `angle APB = 130^(@), and angle BQD = x^(@)`, then the value of x is :

Options

A. 65
B. 130
C. 195
D. 135

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have two circles with centers P and Q that intersect at points B and C. Points A and D lie on the circles with centers P and Q respectively, and points A, C, and D are collinear. We know that \( \angle APB = 130^\circ \) and we need to find \( \angle BQD = x^\circ \).

### Step 2: Draw the Diagram
Draw two intersecting circles with centers P and Q. Mark the points of intersection as B and C. Place point A on the circle centered at P and point D on the circle centered at Q, ensuring that points A, C, and D are collinear.

### Step 3: Use the Inscribed Angle Theorem
According to the Inscribed Angle Theorem, the angle at the center of the circle is twice the angle at the circumference subtended by the same arc.

Let \( \angle APB = 130^\circ \). The angle subtended by arc AB at point C (which we will call \( \angle ACB \)) will be half of \( \angle APB \):
\[ \angle ACB = \frac{1}{2} \times \angle APB = \frac{1}{2} \times 130^\circ = 65^\circ \]

### Step 4: Identify the Cyclic Quadrilateral
Since points A, B, C, and D form a cyclic quadrilateral (as they lie on the circumference of the circles), we can use the property that the sum of the opposite angles in a cyclic quadrilateral is \( 180^\circ \).

Let \( \angle ACB = 65^\circ \) and let \( \angle ADB = z \). According to the cyclic quadrilateral property:
\[ \angle ACB + \angle ADB = 180^\circ \]
Thus,
\[ 65^\circ + z = 180^\circ \]
From this, we can find \( z \):
\[ z = 180^\circ - 65^\circ = 115^\circ \]

### Step 5: Find the Angle at the Center Q
Now, we will find \( \angle BQD \). Since \( \angle BQD \) is the angle at the center Q subtended by the same arc BD, we can use the same theorem again.

Let \( \angle BQD = x \). The angle at the circumference (which is \( \angle ADB = z \)) is:
\[ x = 2 \times z = 2 \times 115^\circ = 230^\circ \]

### Step 6: Final Calculation
However, since angles in a circle must be less than \( 360^\circ \), we need to adjust our angle \( x \):
\[ x = 230^\circ - 180^\circ = 50^\circ \]

### Conclusion
Thus, the value of \( x \) is \( 50^\circ \).