AB and CD are two equal chords of a circle with centre at O. If `OP bot AB` and `OQ bot CD`,where P and Q are points on the chords AB and CD respectively and if `anglePOQ = 100^(@)` , the measure of `angleAPQ` is :

Options

A. `45^(@)`
B. `50^(@)`
C. `60^(@)`
D. `80^(@)`

Explanation

To solve the problem step-by-step, we will analyze the given information and apply geometric principles.

### Step 1: Understand the Configuration
We have a circle with center O, and two equal chords AB and CD. The perpendiculars from O to these chords are OP and OQ, respectively. Points P and Q lie on chords AB and CD.

### Step 2: Given Information
- Chords AB and CD are equal in length.
- OP ⊥ AB and OQ ⊥ CD.
- Angle POQ = 100°.

### Step 3: Establish Relationships
Since AB and CD are equal chords, the distances from the center O to the chords (OP and OQ) must also be equal. Therefore, we can conclude:
\[ OP = OQ \]

### Step 4: Analyze Triangle OPQ
In triangle OPQ, we know:
- Angle POQ = 100°.
- Since OP = OQ, triangle OPQ is isosceles.

Let:
- Angle OPQ = Angle OQP = x (since they are equal).

### Step 5: Use Triangle Sum Property
The sum of angles in triangle OPQ is:
\[ \text{Angle OPQ} + \text{Angle OQP} + \text{Angle POQ} = 180° \]
Substituting the known values:
\[ x + x + 100° = 180° \]
\[ 2x + 100° = 180° \]
\[ 2x = 180° - 100° \]
\[ 2x = 80° \]
\[ x = 40° \]

Thus, we have:
\[ \text{Angle OPQ} = \text{Angle OQP} = 40° \]

### Step 6: Relate to Angle APQ
Now, we need to find angle APQ. We know:
- Angle AOP = 90° (since OP is perpendicular to AB).

Using the angles around point P:
\[ \text{Angle AOP} = \text{Angle APQ} + \text{Angle OPQ} \]
Substituting the known values:
\[ 90° = \text{Angle APQ} + 40° \]

### Step 7: Solve for Angle APQ
Rearranging gives:
\[ \text{Angle APQ} = 90° - 40° \]
\[ \text{Angle APQ} = 50° \]

### Final Answer
The measure of angle APQ is:
\[ \boxed{50°} \]