Given, PA and PB are two tangents.And ∠OPA=35o.
We know that line joining point of contact of tangent to centre of circle is perpendicular to tangent.
∴∠OAP=∠OBP=90∘...(i)
In, △OAP, by angle sum property of a triangle,
∠OAP+∠APO+∠POA=180o
∠AOP=180∘−35∘−90∘
∠AOP=55∘
Consider △OAP and △OBP,
OA=OB (radius )
∠OAP=∠OBP=90∘ (from (i))
PA=PB (Tangents from an external point to the circle are equal in length.)
∴△OAP≅△OBP (By SAS congruence criterion)
By CPCT,∠AOP=∠BOP.
Thus, ∠BOP=55∘.