In `Delta PQR, PD_|_ QR` and PO is the bisector of `/_ QPR`. If `/_ PQR=65^(@)` and `/_ PRQ =23(1)/(2). ^(@)` then `/_ DPO` in degrees `=`

Options

A. `20(3)/(4)`
B. `20(1)/(2)`
C. `20(1)/(5)`
D. `20(1)/(4)`

Explanation

To solve the problem, we need to find the angle \( \angle DPO \) in triangle \( PQR \) given that \( PD \perp QR \) and \( PO \) is the bisector of \( \angle QPR \). We are also given the angles \( \angle PQR = 65^\circ \) and \( \angle PRQ = 23.5^\circ \).

### Step-by-step Solution:

1. **Identify the Angles**:
We know:
\[ \angle PQR = 65^\circ \]
\[ \angle PRQ = 23.5^\circ \]

2. **Find \( \angle QPR \)**:
Using the fact that the sum of angles in a triangle is \( 180^\circ \):
\[ \angle QPR = 180^\circ - \angle PQR - \angle PRQ \]
Substituting the known values:
\[ \angle QPR = 180^\circ - 65^\circ - 23.5^\circ = 180^\circ - 88.5^\circ = 91.5^\circ \]

3. **Determine \( \angle QPO \) and \( \angle OPR \)**:
Since \( PO \) is the bisector of \( \angle QPR \):
\[ \angle QPO = \angle OPR = \frac{1}{2} \angle QPR = \frac{1}{2} \times 91.5^\circ = 45.75^\circ \]

4. **Find \( \angle QPD \)**:
Since \( PD \perp QR \), we can find \( \angle QPD \):
\[ \angle QPD = 180^\circ - \angle PQR - 90^\circ \]
Substituting the known values:
\[ \angle QPD = 180^\circ - 65^\circ - 90^\circ = 25^\circ \]

5. **Calculate \( \angle DPO \)**:
Now we can find \( \angle DPO \) using:
\[ \angle DPO = \angle QPO - \angle QPD \]
Substituting the values we found:
\[ \angle DPO = 45.75^\circ - 25^\circ = 20.75^\circ \]

Thus, the value of \( \angle DPO \) is \( 20.75^\circ \).

### Final Answer:
\[ \angle DPO = 20.75^\circ \]