Given:

_∠OPQ = 35°

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: The sum of all angles of a triangle = 180°.

By property 1,

QP = QR (tangent from Q)

By property 2, ∆OPQ is right-angled at ∠OQP (i.e., ∠OQP = 90°) and ∆ORP is right-angled at ∠ORP (i.e., ∠ORP = 90°).

∴ OQ ⊥ QP

OR⊥ RP

Now,

∠OQP = ∠ORP = 90° [Property 1]

QP = QR [Property 2]

OP = OP [Given]

∴ ∆OPQ ≅ ∆OPR By SAS

Hence, ∠OPQ = ∠OPR = 35° By CPCTC

i.e. ∠a = 35°

By property 3,

∠OQP + ∠OPQ + ∠QOP = 180°

⇒ 90° + 35° + ∠QOP = 180°

⇒ 125° + ∠QOP = 180°

⇒ ∠QOP = 180° - 125°

⇒ ∠QOP = 55°

i.e. ∠b = 55°

Hence, ∠a = 35° and ∠b = 55°