Question

plz help me with this question ​

Answer 1

Answer:

hope it helps

Step-by-step explanation:

InquadrilateralQRP

⇒∠OQP+∠QRP+∠PRO+∠ROQ=360

⇒90+50 +90 +∠ROQ=360

∠ROQ=360 −90 −90 −50

⇒∠ROQ=130

⇒∴∠RSQ= 21

∠ROQ∠RSQ= 21×130=65

Answer 2

Answer:

60°

Step-by-step explanation:

Given that: ∠RPQ = 60°

Now, since OR and OQ is radius and PR AND PQ is tangent to the circle, ∠ORP= 90°AND ∠OQP=90°.

In quadrilateral PROQ,  sum of all angles=360°

∴∠ORP+∠OQP+∠RPQ+∠ROQ=360°

∠ROQ= 120°

Now, angle made by an arc at the center of a circle is double the angle made by it at arc of the circle

∴∠ROQ= 2∠RSQ

∠RSQ= 120/2= 60°