Intervals Containing Value of Expression in Quadrilateral

To find the interval(s) containing the value of $E = 4\alpha\beta\sin\theta^\circ$, let us analyze the given quadrilateral $PQRS$ step-by-step using trigonometric properties and geometry.

Step 1: Angles in $\triangle QRS$

We are given:

• $QR = 1$
• $\angle PQR = 70^\circ$ and $\angle QRS = 70^\circ$
• $\angle PQS = 15^\circ$
• $\angle PRS = 40^\circ$

From the given angles: $\angle SQR = \angle PQR - \angle PQS = 70^\circ - 15^\circ = 55^\circ$ $\angle PRQ = \angle QRS - \angle PRS = 70^\circ - 40^\circ = 30^\circ$

In $\triangle QRS$, the sum of angles is $180^\circ$: $\angle QSR = 180^\circ - (\angle SQR + \angle QRS) = 180^\circ - (55^\circ + 70^\circ) = 55^\circ$

Since $\angle SQR = \angle QSR = 55^\circ$, $\triangle QRS$ is an isosceles triangle with: $RS = QR = 1$

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Step 2: Apply Sine Rule in $\triangle PQR$

In $\triangle PQR$, the angles are:

• $\angle PQR = 70^\circ$
• $\angle PRQ = 30^\circ$
• $\angle QPR = 180^\circ - (70^\circ + 30^\circ) = 80^\circ$

By the Sine Rule in $\triangle PQR$: $\frac{PQ}{\sin(\angle PRQ)} = \frac{QR}{\sin(\angle QPR)}$

Given $PQ = \alpha$ and $QR = 1$: $\frac{\alpha}{\sin 30^\circ} = \frac{1}{\sin 80^\circ} \implies \alpha = \frac{\sin 30^\circ}{\sin 80^\circ} = \frac{1}{2\cos 10^\circ}$

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Step 3: Apply Sine Rule in $\triangle PRS$

In $\triangle PRS$:

• $\angle PRS = 40^\circ$
• $\angle RPS = \theta^\circ$
• $PS = \beta$
• $RS = 1$

By the Sine Rule in $\triangle PRS$: $\frac{PS}{\sin(\angle PRS)} = \frac{RS}{\sin(\angle RPS)}$ $\frac{\beta}{\sin 40^\circ} = \frac{1}{\sin \theta^\circ} \implies \beta \sin \theta^\circ = \sin 40^\circ$

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Step 4: Evaluate the Expression $E = 4\alpha\beta\sin\theta^\circ$

Substituting the expressions for $\alpha$ and $\beta\sin\theta^\circ$: $E = 4 \left( \frac{1}{2\cos 10^\circ} \right) \sin 40^\circ = \frac{2\sin 40^\circ}{\cos 10^\circ}$

We can simplify $E$ using trigonometric identities: $E = \frac{2\sin(30^\circ + 10^\circ)}{\cos 10^\circ} = \frac{2(\sin 30^\circ \cos 10^\circ + \cos 30^\circ \sin 10^\circ)}{\cos 10^\circ} = 1 + \sqrt{3}\tan 10^\circ$

To determine the exact range of $E$, calculate $E^2$: $E^2 = \frac{4\sin^2 40^\circ}{\cos^2 10^\circ} = \frac{2(1 - \cos 80^\circ)}{\cos^2 10^\circ} = \frac{2(1 - \sin 10^\circ)}{1 - \sin^2 10^\circ} = \frac{2}{1 + \sin 10^\circ}$

Since $0 < \sin 10^\circ < \sin 30^\circ = \frac{1}{2}$:

1. $1 + \sin 10^\circ > 1 \implies E^2 < 2 \implies E < \sqrt{2}$
2. $1 + \sin 10^\circ < \frac{3}{2} \implies E^2 > \frac{4}{3} > 1 \implies E > 1$

Hence, the value of $E$ lies in the range: $1 < E < \sqrt{2}$

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Conclusion

Since $E \in (1, \sqrt{2})$:

• $E \in (0, \sqrt{2})$ (Option A is correct)
• $E \in (1, 2)$ (Option B is correct)

Correct Options: (A) and (B)