Let PQRS be a quadrilateral in a plane, where QR = 1, ∠ PQR = ∠ QRS...

Given,

$PQRS$ be a quadrilateral in a plane, where $QR = 1, ∠ PQR = ∠ QRS = 70 ° , ∠ PQS = 15 °$ and $∠ PRS = 40 °$. If $∠ RPS = \theta ° , PQ = \alpha$ and $PS = \beta$,

Now plotting the diagram of given values we get,

Applying sine rule in $∆ PQR$ we get,

$\dfrac{\alpha}{\sin 30 °} = \dfrac{1}{\sin 80 °}$

$⇒ \alpha = \dfrac{1}{2\sin 80 °} . . . \left(i\right)$

Applying sine rule in $∆ PRS$,

$\dfrac{\beta}{sin{40}^{2}} = \dfrac{1}{sin\theta}$

$⇒ \beta sin\theta = \sin 40 ° . . . \left(ii\right)$

From (i) & (ii),

$4\alpha\beta sin\theta = 4 \cdot \dfrac{1}{2\sin 80 °} \cdot \sin 40 ° = \dfrac{1}{\cos 40 °}$

Using $\cos 0 ° = 1, \cos 30 ° = \dfrac{\sqrt{3}}{2}, \cos 45 ° = \dfrac{1}{\sqrt{2}}$ and $\cos 60 ° = \dfrac{1}{2}$

Only Options (A) and (B) are correct.