In the given figure, O is centre of the circle. Circle has 3 tangents.

SOLUTION

We draw OM & ON from centre O
In ⎕ OMPN,
∠ OMP = ∠ ONP = 90^O, ∠ MPN = 45^O
∠ MON = 360 – (180 + 45) = 135^O
As OP is the bisector of ∠ MON , So ∠ MOP(θ + Φ) = ∠NOP(θ + Φ)
Now as OQ is also a bisector (because MQ and QR are tangents)
∠MOQ = ∠QOP and ∠NOR = ∠ROP
therefore, θ = Φ 
hence, 4θ = 135 and
∠ QOR = ½ ∠ MON = 2θ =  ½ ∠ 135^O = 67.5^O