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Question

In $\Delta\text{PQR}$, $\text{QR}$ is extended up to $\text{S}$ so that $\text{RS} = \text{RP}$. If $\angle\text{PRQ} = 70^\circ$ and $\angle\text{QPS} = 110^\circ$ then find the measure of $\angle\text{PQS}$.

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$35^\circ$

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

  1. What is the foot of the altitude from the vertex A of the triangle ABC?

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