PQRST is cyclic pentagon and PT is a diameter, then `angle (PQR) + angle RST` is equal to

Options

A. `180^@`
B. `270^@`
C. `216^@`
D. `144^@`

Answer

To solve the problem, we need to find the sum of angles \( \angle PQR \) and \( \angle RST \) in the cyclic pentagon \( PQRST \) where \( PT \) is a diameter.

### Step-by-Step Solution:

1. **Understanding the Cyclic Pentagon**:
A cyclic pentagon is a five-sided figure where all vertices lie on a single circle. The angles subtended by the same arc are equal.

2. **Using the Property of Cyclic Quadrilaterals**:
Since \( PT \) is a diameter, by the inscribed angle theorem, the angles subtended by the diameter at any point on the circle are right angles. Therefore, \( \angle PQR \) and \( \angle RST \) can be analyzed using this property.

3. **Finding the Sum of Interior Angles of the Pentagon**:
The sum of the interior angles of a pentagon is given by the formula:
\[ \text{Sum of interior angles} = (n - 2) \times 180^\circ \]
where \( n \) is the number of sides. For a pentagon, \( n = 5 \):
\[ \text{Sum} = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ \]

4. **Analyzing the Angles**:
Since \( PT \) is a diameter, we know:
\[ \angle PQR + \angle RST = 90^\circ + 90^\circ = 180^\circ \]
However, we need to find the specific values of \( \angle PQR \) and \( \angle RST \).

5. **Using the Equal Angles Property**:
In a cyclic pentagon, if we assume that the angles \( \angle PQR \) and \( \angle RST \) are equal due to symmetry (as \( PQ = QR = RS = ST \)), we can denote:
\[ \angle PQR = x \quad \text{and} \quad \angle RST = x \]
Therefore, we have:
\[ x + x = 180^\circ \implies 2x = 180^\circ \implies x = 90^\circ \]

6. **Final Calculation**:
Thus, the sum of the angles \( \angle PQR + \angle RST \) is:
\[ \angle PQR + \angle RST = 90^\circ + 90^\circ = 180^\circ \]

### Conclusion:
The sum of the angles \( \angle PQR + \angle RST \) is \( 180^\circ \).