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

Label the unknown lengths as $$PQ=\alpha,\; PR=x,\; QS=y,\; RS=z,\; PS=\beta$$ and the required angle as $$\theta=\angle RPS$$.

Step 1: Fix the angles inside the quadrilateral
Interior angles at the vertices $$Q$$ and $$R$$ are each $$70^\circ$$.
The diagonal $$QS$$ divides $$\angle PQR=70^\circ$$ into $$\angle PQS =15^\circ$$ and $$\angle SQR =70^\circ-15^\circ =55^\circ$$.
Similarly, the diagonal $$PR$$ divides $$\angle QRS=70^\circ$$ into $$\angle PRS =40^\circ$$ and $$\angle QRP =70^\circ-40^\circ =30^\circ$$.

Step 2: Triangle $$\triangle PQR$$
In $$\triangle PQR$$ the three angles are $$\angle PQR =70^\circ,\; \angle QRP =30^\circ,\; \angle QPR =180^\circ-70^\circ-30^\circ =80^\circ.$$
With $$QR=1$$, the sine rule gives $$\frac{\alpha}{\sin30^\circ}=\frac{1}{\sin80^\circ}\;\;\Longrightarrow\;\; \alpha=\frac{\sin30^\circ}{\sin80^\circ}=\frac12\cdot\frac1{\sin80^\circ}.$$

Step 3: Triangle $$\triangle QRS$$
For $$\triangle QRS$$ we already know $$\angle QRS =70^\circ,\; \angle SQR =55^\circ,\; \angle QSR =55^\circ.$$ Hence it is isosceles with $$QR=RS$$, so $$z=RS=QR=1.$$ Applying the sine rule once more gives $$y=QS=\frac{\sin70^\circ}{\sin55^\circ}.$$

Step 4: Triangle $$\triangle PRS$$ and evaluation of $$\sin\theta$$
In $$\triangle PRS$$ the known data are $$PR=x\;(\text{from Step 2}),\; RS=z=1,\; \angle PRS =40^\circ.$$ By the sine rule $$\frac{1}{\sin\theta}=\frac{\beta}{\sin40^\circ}\;\;\Longrightarrow\;\; \sin\theta=\frac{\sin40^\circ}{\beta}.$$

Step 5: Eliminate $$\beta$$ and simplify the required product
Consider the product $$4\alpha\beta\sin\theta=4\alpha\beta\cdot\frac{\sin40^\circ}{\beta}=4\alpha\sin40^\circ.$$ Thus $$\beta$$ and $$\theta$$ disappear - only $$\alpha$$ is still needed.

Step 6: Substitute $$\alpha$$
Using $$\alpha=\dfrac{\sin30^\circ}{\sin80^\circ}$$ obtained in Step 2, $$4\alpha\sin40^\circ =4\left(\frac{\sin30^\circ}{\sin80^\circ}\right)\sin40^\circ =2\cdot\frac{\sin40^\circ}{\sin80^\circ}.$$

Step 7: Recognise the exact value
Since $$\sin80^\circ=2\sin40^\circ\cos40^\circ,$$ $$2\cdot\frac{\sin40^\circ}{\sin80^\circ} =2\cdot\frac{\sin40^\circ}{2\sin40^\circ\cos40^\circ} =\frac1{\cos40^\circ} =\sec40^\circ.$$ Numerically, $$\sec40^\circ\approx1.305.$$

Step 8: Locate the value in the given intervals
$$\sec40^\circ$$ lies between $$1$$ and $$\sqrt2\;( \sqrt2\approx1.414 ).$$
Hence   • it belongs to $$(0,\sqrt{2})$$ (Option A), and
  • it also belongs to $$(1,2)$$ (Option B).

Therefore the correct intervals are:
Option A $$(0,\sqrt{2})$$ and Option B $$(1,2).$$