The correct option is B 50°
Here,∠P+∠Q=180∘ [Sum of co−interior angles is 180∘]⇒∠P+130∘=180∘⇒∠P=180∘−130∘⇒∠P=50∘∠R=90∘ [Given]∴∠S+90∘=180∘[Sum of co−interior angles is 180∘]⇒∠S=180∘−90∘⇒∠S=90∘
Yes, we can use another method to find ∠P∠S+∠R+∠Q+∠P=360∘ [Angle sum property of quadrilateral]⇒90∘+90∘+130∘+∠P=360∘⇒310∘+∠P=360∘⇒∠P=360∘−310∘⇒∠P=50∘