Let PQRS be a quadrilateral in a plane, where QR = 1, PQR = QRS = 70°, PQS = 15° and PRS = 40°. If RPS = °, PQ = and PS = , then the interval(s) that contain(s) the value of 4 sin ° is/are

The correct options are (0, 2) and (1, 2).

Step 1: Determine the angles and sides in triangle QSR

From the given angle measurements in the quadrilateral PQRS:

$\angle \text{SQR}=\angle \text{PQR}-\angle \text{PQS}={70}^{\circ }-{15}^{\circ }={55}^{\circ }$

$\angle \text{QRS}={70}^{\circ }$

In △QSR, the sum of angles is 180°:

$\angle \text{QSR}={180}^{\circ }-({55}^{\circ }+{70}^{\circ })={55}^{\circ }$

Since $\angle \text{SQR}=\angle \text{QSR}={55}^{\circ }$, △QSR is an isosceles triangle with:

$\text{RS}=\text{QR}=1$

Step 2: Determine the angles and sides in triangle PQR

In △PQR:

$\angle \text{PRQ}=\angle \text{QRS}-\angle \text{PRS}={70}^{\circ }-{40}^{\circ }={30}^{\circ }$

$\angle \text{PQR}={70}^{\circ }$

$\angle \text{QPR}={180}^{\circ }-({70}^{\circ }+{30}^{\circ })={80}^{\circ }$

Applying the Law of Sines in △PQR:

$\frac{\text{PR}}{sin{70}^{\circ }}=\frac{\text{QR}}{sin{80}^{\circ }}$

Since QR = 1, we get:

$\text{PR}=\frac{sin{70}^{\circ }}{sin{80}^{\circ }}$

Step 3: Calculate angle α in triangle PRS

In △PRS, we have $\angle \text{PRS}={40}^{\circ }$ and $\angle \text{RPS}={\alpha }^{\circ }$. Therefore, the third angle is:

$\angle \text{PSR}={180}^{\circ }-({40}^{\circ }+{\alpha }^{\circ })$

Applying the Law of Sines in △PRS:

$\frac{\text{RS}}{sin{\alpha }^{\circ }}=\frac{\text{PR}}{sin\angle \text{PSR}}=\frac{\text{PR}}{sin({40}^{\circ }+{\alpha }^{\circ })}$

Substituting RS = 1 and $\text{PR}=\frac{sin{70}^{\circ }}{sin{80}^{\circ }}$:

$\frac{sin({40}^{\circ }+{\alpha }^{\circ })}{sin{\alpha }^{\circ }}=\frac{sin{70}^{\circ }}{sin{80}^{\circ }}$

Testing $\alpha =15$:

$\frac{sin{55}^{\circ }}{sin{15}^{\circ }}=\frac{sin{70}^{\circ }}{sin{80}^{\circ }}$

Cross-multiplying gives $sin{55}^{\circ }sin{80}^{\circ }=sin{70}^{\circ }sin{15}^{\circ }$, which holds true identically. Thus, $\alpha =15$.

Step 4: Compute 4 sin α° and evaluate intervals

For $\alpha =15$:

$4sin{15}^{\circ }=4\times \frac{\sqrt{6}-\sqrt{2}}{4}=\sqrt{6}-\sqrt{2}$

Approximating the value:

$\sqrt{6}-\sqrt{2}\approx 2.449-1.414=1.035$

Since 1.035 lies inside both (0, 2) and (1, 2), the correct options are (0, 2) and (1, 2).