S is a point on side PQ of a triangle PQR such that RS = PS = QS. If $\angle \text{QPR}={15}^{\circ }$, then the value of (2 $\angle \text{PQR}$ − $\angle \text{PRQ}$) is :

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The correct option is 60°.

Given that $S$ is a point on side $PQ$ of $△PQR$ such that $RS=PS=QS$.

Also given that $\angle \text{QPR}={15}^{\circ }$.

Step 1: Use properties of isosceles triangles

In $△PSR$, since $PS=RS$, the angles opposite to these sides are equal:

$\angle \text{PRS}=\angle \text{QPR}={15}^{\circ }$

In $△QSR$, since $QS=RS$, the angles opposite to these sides are equal:

$\angle \text{QRS}=\angle \text{PQR}$

Therefore, the total angle at vertex $R$ is:

$\angle \text{PRQ}=\angle \text{PRS}+\angle \text{QRS}={15}^{\circ }+\angle \text{PQR}$

Step 2: Apply angle sum property of triangle

In $△PQR$, the sum of all interior angles is ${180}^{\circ }$:

$\angle \text{QPR}+\angle \text{PQR}+\angle \text{PRQ}={180}^{\circ }$

Substituting the values:

${15}^{\circ }+\angle \text{PQR}+({15}^{\circ }+\angle \text{PQR})={180}^{\circ }$

${30}^{\circ }+2\angle \text{PQR}={180}^{\circ }$

$2\angle \text{PQR}={150}^{\circ }$

$\angle \text{PQR}={75}^{\circ }$

Now, calculate $\angle \text{PRQ}$:

$\angle \text{PRQ}={15}^{\circ }+{75}^{\circ }={90}^{\circ }$

Step 3: Calculate the required value

We need to find the value of $(2\angle \text{PQR}-\angle \text{PRQ})$:

$2\angle \text{PQR}-\angle \text{PRQ}=2\left({75}^{\circ }\right)-{90}^{\circ }$

$2\angle \text{PQR}-\angle \text{PRQ}={150}^{\circ }-{90}^{\circ }={60}^{\circ }$

Thus, the required value is 60°.

``` Let's double check constraints and instructions: - Output format: MUST output BOTH `` and `` tags. - No markdown block ticks or conversational text outside tags. Everything looks completely precise and compliant. 60°

The correct answer is 60°.

Given that $S$ is a point on side $PQ$ of $△PQR$ such that $RS=PS=QS$.

Also given that $\angle \text{QPR}={15}^{\circ }$.

Step 1: Use properties of isosceles triangles

In $△PSR$, since $PS=RS$, the angles opposite to these equal sides are equal:

$\angle \text{PRS}=\angle \text{QPR}={15}^{\circ }$

In $△QSR$, since $QS=RS$, the angles opposite to these equal sides are equal:

$\angle \text{QRS}=\angle \text{PQR}$

Therefore, the total angle at vertex $R$ is:

$\angle \text{PRQ}=\angle \text{PRS}+\angle \text{QRS}={15}^{\circ }+\angle \text{PQR}$

Step 2: Apply angle sum property of triangle PQR

In $△PQR$, the sum of all interior angles is ${180}^{\circ }$:

$\angle \text{QPR}+\angle \text{PQR}+\angle \text{PRQ}={180}^{\circ }$

Substituting $\angle \text{QPR}={15}^{\circ }$ and $\angle \text{PRQ}={15}^{\circ }+\angle \text{PQR}$:

${15}^{\circ }+\angle \text{PQR}+({15}^{\circ }+\angle \text{PQR})={180}^{\circ }$

${30}^{\circ }+2\angle \text{PQR}={180}^{\circ }$

$2\angle \text{PQR}={150}^{\circ }$

$\angle \text{PQR}={75}^{\circ }$

Now, calculate $\angle \text{PRQ}$:

$\angle \text{PRQ}={15}^{\circ }+{75}^{\circ }={90}^{\circ }$

Step 3: Find the value of (2 ∠PQR − ∠PRQ)

$2\angle \text{PQR}-\angle \text{PRQ}=2\left({75}^{\circ }\right)-{90}^{\circ }$

$2\angle \text{PQR}-\angle \text{PRQ}={150}^{\circ }-{90}^{\circ }={60}^{\circ }$

Thus, the required value is 60°.