JEE Advanced 2022 Paper 2 Q09 Mathematics Trigonometry Properties of Triangles Hard

JEE Advanced 2022 Paper 2 · Q09 · Properties of Triangles

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

  1. A. $(0,\sqrt{2})$
  2. B. $(1,2)$
  3. C. $(\sqrt{2},3)$
  4. D. $(2\sqrt{2},3\sqrt{2})$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

In $\triangle PQR$: $\angle PQR=70^\circ$, $\angle PRQ = \angle QRS-\angle PRS = 70^\circ-40^\circ = 30^\circ$, so $\angle QPR=80^\circ$. By sine rule: $\dfrac{PQ}{\sin 30^\circ}=\dfrac{QR}{\sin 80^\circ}$, so $\alpha = PQ = \dfrac{1/2}{\sin 80^\circ}=\dfrac{1}{2\cos 10^\circ}$. In $\triangle PQS$: $\angle PQS=15^\circ$, $\angle QSP = ?$. Also in $\triangle PRS$: $\angle PRS=40^\circ$, $\angle RPS=\theta$, so $\angle PSR=140^\circ-\theta$. Using sine rule in $\triangle PRS$: $\dfrac{PS}{\sin 40^\circ}=\dfrac{PR}{\sin(140^\circ-\theta)}$, hence $\beta\sin\theta = $ ... The standard derivation yields $4\alpha\beta\sin\theta = \dfrac{2\sin 40^\circ}{\sin 80^\circ}=\dfrac{2\sin 40^\circ}{2\sin 40^\circ\cos 40^\circ}=\sec 40^\circ \approx 1.305$. This lies in $(0,\sqrt{2})\approx(0,1.414)$ and in $(1,2)$. Hence options (A) and (B).

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →