PQ is a diameter of a circle with centre O. RS is a chord parallel to PQ that subtends an angle of `40^(@)` at the centre of the cirle. If PR and QS are produced to meet at T, Then what will be the measure (in degrees) of `/_PTQ`?

Options

A. 55
B. 60
C. 70
D. 90

Answer

To solve the problem, we need to find the measure of angle \( \angle PTQ \) given the conditions of the circle and the angles formed by the chords and diameters. Here’s a step-by-step solution:

### Step 1: Understand the Circle and Angles
- We have a circle with center \( O \) and diameter \( PQ \).
- The chord \( RS \) is parallel to \( PQ \) and subtends an angle of \( 40^\circ \) at the center \( O \).

### Step 2: Apply the Angle at the Center Theorem
- According to the theorem, the angle subtended at the center of the circle is twice the angle subtended at the circumference.
- Therefore, if \( \angle ROS = 40^\circ \) (the angle at the center), then the angle at the circumference \( \angle RQS \) (formed by the endpoints of the chord \( RS \)) will be:
\[ \angle RQS = \frac{1}{2} \times \angle ROS = \frac{1}{2} \times 40^\circ = 20^\circ \]

### Step 3: Determine Other Angles in the Circle
- Since \( PQ \) is the diameter, it subtends a right angle at any point on the circumference. Thus:
\[ \angle PSQ = 90^\circ \]

### Step 4: Use the Exterior Angle Theorem
- In triangle \( PST \), we can apply the exterior angle theorem, which states that the exterior angle is equal to the sum of the two opposite interior angles.
- Here, \( \angle PSQ \) is the exterior angle for triangle \( PST \):
\[ \angle PSQ = \angle SPT + \angle PTS \]
- Substituting the known values:
\[ 90^\circ = 20^\circ + \angle PTS \]

### Step 5: Solve for \( \angle PTS \)
- Rearranging the equation gives:
\[ \angle PTS = 90^\circ - 20^\circ = 70^\circ \]

### Step 6: Conclusion
- Since \( \angle PTQ \) is equal to \( \angle PTS \), we have:
\[ \angle PTQ = 70^\circ \]

Thus, the measure of \( \angle PTQ \) is \( 70^\circ \).