(b) 45o
From the given figure,
In triangle PQS, ∠PSQ + ∠QPS = 110o … [from exterior angle property of a triangle]
We know that, sum of all angle of the triangle is equal to 180o.
So, ∠PSQ + ∠QPS + ∠PQS = 180o
110o + ∠PQS = 180o – 110o
∠PQS = 70o
Now, consider the triangle PRS,
∠PSQ = x + 25o … [from the exterior angle property of a triangle]
x = 70o – 25o
x = 45o