Answer:
[tex]\sf \angle RPS = \boxed{46^\circ}[/tex]
Step-by-step explanation:
Given
- [tex]\sf P[/tex], [tex]\sf Q[/tex], [tex]\sf R[/tex], and [tex]\sf S[/tex] are points on a circle with center [tex]\sf O[/tex], and
- [tex]\sf PS[/tex] is a diameter of the circle.
- [tex]\sf \angle PQR = 136^\circ[/tex].
To find:
Size of angle [tex]\sf RPS[/tex]
Solution:
Identify Cyclic Quadrilateral Property:
Recognize that [tex]\sf PQRS[/tex] is a cyclic quadrilateral. In a cyclic quadrilateral, opposite angles are supplementary (they sum up to [tex]\sf 180^\circ[/tex]).
Apply Opposite Angle Property:
We know that [tex]\sf \angle PQR[/tex] and [tex]\sf \angle PSR[/tex] are opposite angles in the cyclic quadrilateral [tex]\sf PQRS[/tex]. Therefore:
[tex]\sf \angle PSR + \angle PQR = 180^\circ [/tex]
Hence,
[tex]\sf \angle PSR + 136^\circ = 180^\circ [/tex]
Solving for [tex]\sf \angle PSR[/tex]:
[tex]\sf \angle PSR = 180^\circ - 136^\circ [/tex]
[tex]\sf \angle PSR = 44^\circ [/tex]
Use Angle in a Semicircle:
Given that [tex]\sf PS[/tex] is a diameter of the circle, [tex]\sf \angle PRS = 90^\circ[/tex] (an angle in a semicircle).
Apply Triangle Angle Sum Property:
In triangle [tex]\sf PRS[/tex], the sum of the angles is [tex]\sf 180^\circ[/tex]. Therefore:
[tex]\sf \angle RPS + \angle PRS + \angle PSR = 180^\circ [/tex]
Substituting the known values:
[tex]\sf \angle RPS + 90^\circ + 44^\circ = 180^\circ [/tex]
Simplifying:
[tex]\sf \angle RPS + 134^\circ = 180^\circ [/tex]
Solving for [tex]\sf \angle RPS[/tex]:
[tex]\sf \angle RPS = 180^\circ - 134^\circ [/tex]
[tex]\sf \angle RPS = 46^\circ [/tex]
Conclusion:
Therefore, the size of angle [tex]\sf RPS[/tex] is [tex]\sf \boxed{46^\circ}[/tex].
