Short Answer 
The problem involves angles ‚a†PQR (45¬∞), ‚a†POR (90¬∞), ‚a†OPQ (25¬∞), and ‚a†ORQ (20¬∞) in a circle with center O. Using properties of isosceles triangles, it was determined that ‚a†OQP equals 25¬∞ and ‚a†OQR equals 20¬∞, leading to the confirmation that ‚a†PQR is 45¬∞ and ‚a†POR is 90¬∞.
Step 1: Identify Given Angles
The problem provides the following angles which are critical for our calculations:
- ‚a†PQR = 45¬∞
- ‚a†POR = 90¬∞
- ‚a†OPQ = 25¬∞
- ‚a†ORQ = 20¬∞
It is important to note that O is the center of the circle, and the radius OP is equal to OQ.
Step 2: Apply Properties of Circles
Using the properties of isosceles triangles, recognize that:
- OP = OQ (radii of the circle)
- Thus, ‚a†OPQ = ‚a†OQP = 25¬∞.
We can also find ‚a†OQR by realizing that it equals ‚a†ORQ, which is given as 20¬∞.
Step 3: Calculate the Required Angles
Now that we have all necessary angles, we can find the required angles:
- To find ‚a†PQR: ‚a†PQR = ‚a†OQP + ‚a†OQR = 25¬∞ + 20¬∞ = 45¬∞.
- To find ‚a†POR: Since it is twice ‚a†PQR, ‚a†POR = 2 * ‚a†PQR = 2 * 45¬∞ = 90¬∞.
Thus, the final angle values are: ‚a†PQR = 45¬∞ and ‚a†POR = 90¬∞.