Question 9 - Jee advanced Math 2022 P2 Questions with Solutions

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 \(4\alpha \beta \sin{\theta}\) is/are

A) \((0, \sqrt{2})\)

B) \((1, 2)\)

C) \((\sqrt{2}, 3)\)

D) \((2\sqrt{2}, 3\sqrt{2})\)

Sol : 

The given information is coded in the following figure :

\(\angle{SQR} = 70^{\circ} - 15^{\circ} = 55^{\circ}\)

\(\angle{PRQ} = 70^{\circ} - 40^{\circ} = 30^{\circ}\)

In triangle \(QSR\),

\(\angle{QSR} = 180^{\circ} - (55^{\circ} + (40^{\circ} + 30^{\circ}))\) 

  \(= 55^{\circ}= \angle{SQR}\)

\(\implies QR = SR = 1\)….{in a triangle, sides opposite to equal angles are equal}

In triangle \(PQR\),

\(\angle{QPR} = 180^{\circ} - (30^{\circ} + (15^{\circ} + 55^{\circ}))\) 

  \(= 80^{\circ}\)

Using Sine rule of triangles,

\(\frac{\alpha}{\sin{30^{\circ}}} = \frac{1}{\sin{80^{\circ}}}\)

\(\implies \alpha = \frac{\sin{30^{\circ}}}{\sin{80^{\circ}}}\)


In triangle \(PRS\),

Using Sine rule of triangles,

\(\frac{\beta}{\sin{40^{\circ}}} = \frac{1}{\sin{\theta}}\)

\(\implies \beta \sin{\theta} = \sin{40^{\circ}}\)

So,

\(4 \alpha \beta \sin{\theta}= 4 (\frac{\sin{30^{\circ}}}{\sin{80^{\circ}}}) \sin{40^{\circ}}\)

             \(= 4 (\frac{\frac{1}{2}}{2\sin{40^{\circ}} \cos{40^{\circ}}}) \sin{40^{\circ}}\)

              \(= \frac{1}{\cos{40^{\circ}}}\)

\(\cos{45^{\circ}} < \cos{40^{\circ}} < \cos{30^{\circ}}\)

\(\frac{1}{\sqrt{2}} < \cos{40^{\circ}} < \frac{\sqrt{3}}{2}\)

\(\frac{2}{\sqrt{3}} < \frac{1}{\cos{40^{\circ}}} < \frac{\sqrt{2}}{1}\)

\(\implies \frac{1}{\cos{40^{\circ}}} \in (\frac{2}{\sqrt{3}}, \frac{\sqrt{2}}{1})\)

      \(\approx (1.01, 1.41)\)

Therefore, options A and B are True.

 

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