Let PQRS be a quadrilateral in a plane,where | vedclass

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Question

(C) In $	riangle PQR$,$\angle PRQ = \angle QRS - \angle PRS = 70^{\circ} - 40^{\circ} = 30^{\circ}$.In $	riangle QRS$,$\angle RQS = \angle PQR - \an

Vedclass · Accepted Answer

$(A), (B)$

Vedclass · Answer

(C) In $	riangle PQR$,$\angle PRQ = \angle QRS - \angle PRS = 70^{\circ} - 40^{\circ} = 30^{\circ}$.In $	riangle QRS$,$\angle RQS = \angle PQR - \angle PQS = 70^{\circ} - 15^{\circ} = 55^{\circ}$.Then $\angle QSR = 180^{\circ} - 70^{\circ} - 55^{\circ} = 55^{\circ}$.Since $\angle RQS = \angle QSR = 55^{\circ}$,we have $QR = RS = 1$.In $	riangle PQR$,$\angle QPR = 180^{\circ} - 70^{\circ} - 30^{\circ} = 80^{\circ}$.Applying the sine rule in $	riangle PQR$:$\frac{\alpha}{\sin 30^{\circ}} = \frac{1}{\sin 80^{\circ}} \Rightarrow \alpha = \frac{1}{2 \sin 80^{\circ}}$.Applying the sine rule in $	riangle PRS$:$\frac{\beta}{\sin 40^{\circ}} = \frac{1}{\sin 	heta} \Rightarrow \beta \sin 	heta = \sin 40^{\circ}$.Now,$4 \alpha \beta \sin 	heta = 4 \left( \frac{1}{2 \sin 80^{\circ}} ight) \sin 40^{\circ} = \frac{2 \sin 40^{\circ}}{\sin 80^{\circ}} = \frac{2 \sin 40^{\circ}}{2 \sin 40^{\circ} \cos 40^{\circ}} = \sec 40^{\circ}$.Since $30^{\circ} Since $\frac{2}{\sqrt{3}} \approx 1.15$ and $\sqrt{2} \approx 1.414$,the value $\sec 40^{\circ}$ lies in the intervals $(0, \sqrt{2})$ and $(1, 2)$.Thus,the correct options are $(A)$ and $(B)$.

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