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

(A) (A) $(0, \sqrt{2})$; (B) (B) $(1, 2)$

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

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Properties of Triangle

Let PQRS be a quadrilateral in a plane, where

QR = 1

∠ P QR = ∠ QR S = 7 0^{\circ}

∠ P QS = 1 5^{\circ}

and

∠ P R S = 4 0^{\circ}

. If

∠ R P S = θ^{\circ}

P Q = α

and

P S = β

, then the interval(s) that contain(s) the value of

4 α β sin θ^{\circ}

is/are

Answer: (A), (B)

From the given angles:

Since

∠ R QS = ∠ QS R = 5 5^{\circ}

, triangle QRS is isosceles with

QR = R S = 1

.

Also,

∠ QP R = 18 0^{\circ} - 7 0^{\circ} - 3 0^{\circ} = 8 0^{\circ}

.

Applying sine rule in

△ P R Q

\frac{α}{sin 3 0 ^{\circ}} = \frac{1}{sin 8 0 ^{\circ}} \Rightarrow α = \frac{1}{2 sin 8 0 ^{\circ}}

Applying sine rule in

△ P R S

\frac{β}{sin 4 0 ^{\circ}} = \frac{1}{sin θ} \Rightarrow β sin θ = sin 4 0^{\circ}

Therefore:

4 α β sin θ = \frac{4 sin 4 0 ^{\circ}}{2 sin 8 0 ^{\circ}} = \frac{4 sin 4 0 ^{\circ}}{4 sin 4 0 ^{\circ} cos 4 0 ^{\circ}} = sec 4 0^{\circ}

Since

sec 3 0^{\circ} < sec 4 0^{\circ} < sec 4 5^{\circ}

\frac{2}{3} < sec 4 0^{\circ} < 2

1.15 < sec 4 0^{\circ} < 1.414

This lies in intervals (A) and (B).

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · One or more correct options