Given information:
- In $\Delta\text{PQR}$, $\angle\text{PRQ} = 70^\circ$.
- QR is extended to S, making Q-R-S a straight line segment.
- RS = RP, which implies $\Delta\text{PRS}$ is an isosceles triangle.
- $\angle\text{QPS} = 110^\circ$.
- Goal: Find the measure of $\angle\text{PQS}$.
Determine Angle PRS
Since QRS forms a straight line, the angle $\angle\text{PRS}$ is the angle adjacent to $\angle\text{PRQ}$ along the straight line formed by extending QR beyond R. Therefore, $\angle\text{PRS}$ and $\angle\text{PRQ}$ are supplementary if P was on the line, but here $\angle\text{PRS}$ is the angle inside $\Delta\text{PRS}$ adjacent to the line segment QR.
The angle $\angle\text{PRS}$ is calculated as $180^\circ - \angle\text{PRQ}$ because S lies on the extension of QR beyond R.
$\angle\text{PRS} = 180^\circ - \angle\text{PRQ} = 180^\circ - 70^\circ = 110^\circ$.
Solve Isosceles Triangle PRS
In $\Delta\text{PRS}$, since RS = RP, the angles opposite these equal sides are equal:
$\angle\text{RSP} = \angle\text{RPS}$. Let this common angle measure be $x$.
The sum of angles in $\Delta\text{PRS}$ must be $180^\circ$:
$\angle\text{PRS} + \angle\text{RSP} + \angle\text{RPS} = 180^\circ$
$110^\circ + x + x = 180^\circ$
$110^\circ + 2x = 180^\circ$
$2x = 180^\circ - 110^\circ$
$2x = 70^\circ$
$x = \frac{70^\circ}{2} = 35^\circ$.
Identify Angle PQS
Since Q, R, and S are collinear points in the order Q-R-S, the angle $\angle\text{PQS}$ is the same angle as $\angle\text{RSP}$.
Therefore, $\angle\text{PQS} = \angle\text{RSP} = x = 35^\circ$.
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Important Questions from Triangles
What is the circumcenter of the triangle ABC?
What is the centroid of the triangle ABC?
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
