It is given that,
∠SPR = 135 and ∠PQT = 110
∠SPR + ∠QPR = 180 (Linear pair angles)
135 + ∠QPR = 180
∠QPR = 45
Also, ∠PQT + ∠PQR = 180 (Linear pair angles)
110 + ∠PQR = 180
∠PQR = 70
As the sum of all interior angles of a triangle is 180, therefore, for ΔPQR,
∠QPR + ∠PQR + ∠PRQ = 180
45 + 70 + ∠PRQ = 180
∠PRQ = 180 − 115
∠PRQ = 65