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 circle. If PR and QS are produced to meet at T, then what will be the mea sure (in degrees) of `/_PTS`?

Options

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

Answer

To solve the problem, let's follow these steps:

### Step 1: Understand the Given Information
We have a circle with diameter PQ and center O. RS is a chord that is parallel to PQ and subtends an angle of 40 degrees at the center O.

### Step 2: Identify Angles in Triangle ROS
In triangle ROS:
- We know that the angle ROS = 40 degrees.
- Since OR and OS are radii of the circle, we have OR = OS.

### Step 3: Use the Angle Sum Property of Triangle
The sum of angles in triangle ROS is 180 degrees. Therefore, we can write:
\[ \angle ROS + \angle ORS + \angle OSR = 180^\circ \]
Let \( \angle ORS = \angle OSR = x \). Thus, we have:
\[ 40 + x + x = 180 \]
\[ 40 + 2x = 180 \]
\[ 2x = 140 \]
\[ x = 70 \]
So, \( \angle ORS = \angle OSR = 70^\circ \).

### Step 4: Relate Angles with Diameter
Since PQ is a diameter and RS is parallel to PQ, we can use the property of alternate angles:
- \( \angle POR = \angle ORS = 70^\circ \)
- \( \angle QOS = \angle OSR = 70^\circ \)

### Step 5: Analyze Triangle POR
In triangle POR:
- OR = OP (both are radii)
- Therefore, \( \angle OPR = \angle ORP \).
Let \( \angle OPR = \angle ORP = y \). Using the angle sum property again:
\[ \angle OPR + \angle ORP + \angle POR = 180^\circ \]
\[ y + y + 70 = 180 \]
\[ 2y + 70 = 180 \]
\[ 2y = 110 \]
\[ y = 55 \]
So, \( \angle OPR = \angle ORP = 55^\circ \).

### Step 6: Find Angle PTS
Now, we need to find \( \angle PTS \). Since PR and QS are extended to meet at T, we can use the exterior angle theorem:
\[ \angle PTS = \angle POR + \angle OQS \]
Since \( \angle OQS = 70^\circ \) (as established earlier),
\[ \angle PTS = 70 + 70 = 140^\circ \]

### Step 7: Conclusion
Thus, the measure of \( \angle PTS \) is \( 70^\circ \).