Circum-centre of `DeltaPQR` is `O`. If `/_QPR = 55^@` and `/_QRP = 75^@`, what is the value (in degrees) of `/_OPR`?

Options

A. `45 `
B. `40 `
C. `65 `
D. `70 `

Answer

To find the value of angle \( \angle OPR \) in triangle \( PQR \) with circumcenter \( O \), we can follow these steps:

### Step 1: Identify the given angles
We know:
- \( \angle QPR = 55^\circ \)
- \( \angle QRP = 75^\circ \)

### Step 2: Calculate the third angle \( \angle PQR \)
Using the angle sum property of triangles, which states that the sum of the angles in a triangle is \( 180^\circ \):
\[ \angle PQR = 180^\circ - \angle QPR - \angle QRP \]
Substituting the known values:
\[ \angle PQR = 180^\circ - 55^\circ - 75^\circ \]
\[ \angle PQR = 180^\circ - 130^\circ = 50^\circ \]

### Step 3: Understanding the angles at the circumcenter
In triangle \( POR \), where \( O \) is the circumcenter, the angle \( \angle POR \) at the center is twice the angle \( \angle PQR \) at the circumference:
\[ \angle POR = 2 \times \angle PQR = 2 \times 50^\circ = 100^\circ \]

### Step 4: Set up the equation for triangle \( POR \)
In triangle \( POR \), we have:
- \( \angle OPR = x \)
- \( \angle ORP = x \)
- \( \angle POR = 100^\circ \)

Using the angle sum property again:
\[ \angle OPR + \angle ORP + \angle POR = 180^\circ \]
Substituting the known values:
\[ x + x + 100^\circ = 180^\circ \]
\[ 2x + 100^\circ = 180^\circ \]

### Step 5: Solve for \( x \)
Subtract \( 100^\circ \) from both sides:
\[ 2x = 180^\circ - 100^\circ \]
\[ 2x = 80^\circ \]
Now, divide by 2:
\[ x = 40^\circ \]

### Conclusion
Thus, the value of \( \angle OPR \) is \( 40^\circ \).

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