Exterior Angle Theorem Application

Consider $\Delta PQR$, where QR is extended to S, forming a straight line QRS. The angle $\angle PRS = 110^\circ$ is the exterior angle to $\Delta PQR$ at vertex R.

The exterior angle theorem states that an exterior angle equals the sum of the two opposite interior angles ($\angle PQR$ and $\angle RPQ$).

Calculating $\angle PQR$

Using the exterior angle theorem:

$ \angle PRS = \angle PQR + \angle RPQ $

Substitute the given values $\angle PRS = 110^\circ$ and $\angle RPQ = 55^\circ$:

$ 110^\circ = \angle PQR + 55^\circ $

Solve for $\angle PQR$:

$ \angle PQR = 110^\circ - 55^\circ $

$ \angle PQR = 55^\circ $

Final Result for $\angle PQS$

Since Q, R, S are collinear, $\angle PQS$ is the same angle as $\angle PQR$.

Therefore, $\angle PQS = 55^\circ$.