O and C are respectively orthocentre and circumcentre of a `Delta PQR`. P and O joined and produced to meet side QR at S. `angle PQS=60^(@), angleQCR=130^(@)`. Find `angle RPS`.

Options

A. 30
B. 35
C. 40
D. 45

Answer

To solve the problem step-by-step, we will use the properties of triangles and the relationships between angles formed by the orthocenter and circumcenter.

### Step 1: Understand the Given Information
- We have triangle \( PQR \) with orthocenter \( O \) and circumcenter \( C \).
- \( \angle PQS = 60^\circ \)
- \( \angle QCR = 130^\circ \)

### Step 2: Use the Property of the Circumcenter
Since \( C \) is the circumcenter, the angle \( \angle QCR \) subtended by chord \( QR \) at the circumcenter is twice the angle subtended at any point on the circumference. Thus, we can find \( \angle QPR \):
\[ \angle QPR = \frac{1}{2} \times \angle QCR = \frac{1}{2} \times 130^\circ = 65^\circ \]

### Step 3: Use the Property of the Orthocenter
Since \( O \) is the orthocenter, the angle \( \angle PQR \) can be found using the fact that the angles in triangle \( PQR \) sum up to \( 180^\circ \):
\[ \angle PQR + \angle QRP + \angle QPR = 180^\circ \]
Let \( \angle QRP = x \). Thus,
\[ 60^\circ + x + 65^\circ = 180^\circ \]
\[ x = 180^\circ - 125^\circ = 55^\circ \]
So, \( \angle QRP = 55^\circ \).

### Step 4: Find \( \angle RPS \)
Now, we can find \( \angle RPS \) using the relationship:
\[ \angle RPS = \angle QPR - \angle PQS \]
Substituting the known values:
\[ \angle RPS = 65^\circ - 60^\circ = 5^\circ \]

### Final Answer
Thus, the angle \( \angle RPS \) is \( 5^\circ \).

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