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Trigonometric Functions and Equations question

2022 · Shift 2 · Q27
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  5. /2022 · Shift 2 · Q27

Trigonometric Functions and Equations question

2022 · Shift 2 · Q27

JEE AdvancedMathematicsTrigonometric Functions and EquationsMultiple correct+4 / −2
Let PQRSP Q R SPQRS be a quadrilateral in a plane, where QR=1,∠PQR=∠QRS=70∘,∠PQS=15∘Q R=1, \angle P Q R=\angle Q R S=70^{\circ}, \angle P Q S=15^{\circ}QR=1,∠PQR=∠QRS=70∘,∠PQS=15∘ and ∠PRS=40∘\angle P R S=40^{\circ}∠PRS=40∘. If ∠RPS=θ∘,PQ=α\angle R P S=\theta^{\circ}, P Q=\alpha∠RPS=θ∘,PQ=α and PS=βP S=\betaPS=β, then the interval(s) that contain(s) the value of 4αβsin⁡θ∘4 \alpha \beta \sin \theta^{\circ}4αβsinθ∘ is/are
  1. A
    (0,2)(0, \sqrt{2})(0,2​)
  2. B
    (1,2)(1,2)(1,2)
  3. C
    (2,3)(\sqrt{2}, 3)(2​,3)
  4. D
    (22,32)(2 \sqrt{2}, 3 \sqrt{2})(22​,32​)
View written solutionFree

Correct answer: A, B

  1. Interpret the geometry

We need the value of 4αβsin⁡θ,4\alpha\beta\sin\theta,4αβsinθ, where α=PQ,β=PS,θ=∠RPS.\alpha=PQ,\quad \beta=PS,\quad \theta=\angle RPS.α=PQ,β=PS,θ=∠RPS.

Notice that in triangle PQSPQSPQS, the diagonal PRPRPR splits the quadrilateral into triangles PQRPQRPQR and PRSPRSPRS.

A very useful area identity is: [PRS]=12 PR⋅PSsin⁡∠RPS=12 PR⋅βsin⁡θ.[PRS]=\frac12\,PR\cdot PS\sin\angle RPS=\frac12\,PR\cdot \beta\sin\theta.[PRS]=21​PR⋅PSsin∠RPS=21​PR⋅βsinθ. So if we can find PRPRPR and [PRS][PRS][PRS], then we can get βsin⁡θ\beta\sin\thetaβsinθ.

Also from triangle PQRPQRPQR, [PQR]=12 PQ⋅QRsin⁡70∘=12αsin⁡70∘,[PQR]=\frac12\,PQ\cdot QR\sin70^\circ=\frac12\alpha\sin70^\circ,[PQR]=21​PQ⋅QRsin70∘=21​αsin70∘, since QR=1QR=1QR=1.

  1. Find ∠QPR\angle QPR∠QPR in triangle PQRPQRPQR

Given: ∠PQR=70∘.\angle PQR=70^\circ.∠PQR=70∘. Also, ∠PQS=15∘,\angle PQS=15^\circ,∠PQS=15∘, and PRPRPR lies inside angle QPSQPSQPS in the configuration, while from the data at RRR we have ∠PRS=40∘,∠QRS=70∘.\angle PRS=40^\circ,\quad \angle QRS=70^\circ.∠PRS=40∘,∠QRS=70∘. Hence at RRR in triangle PQRPQRPQR, ∠PRQ=∠QRS−∠PRS=70∘−40∘=30∘.\angle PRQ=\angle QRS-\angle PRS=70^\circ-40^\circ=30^\circ.∠PRQ=∠QRS−∠PRS=70∘−40∘=30∘. Therefore in triangle PQRPQRPQR, ∠QPR=180∘−70∘−30∘=80∘.\angle QPR=180^\circ-70^\circ-30^\circ=80^\circ.∠QPR=180∘−70∘−30∘=80∘.

  1. Use sine rule in triangle PQRPQRPQR to find α=PQ\alpha=PQα=PQ and PRPRPR

In triangle PQRPQRPQR, PQsin⁡30∘=QRsin⁡80∘=PRsin⁡70∘.\frac{PQ}{\sin30^\circ}=\frac{QR}{\sin80^\circ}=\frac{PR}{\sin70^\circ}.sin30∘PQ​=sin80∘QR​=sin70∘PR​. Since QR=1QR=1QR=1, PQ=sin⁡30∘sin⁡80∘=12sin⁡80∘.PQ=\frac{\sin30^\circ}{\sin80^\circ}=\frac{1}{2\sin80^\circ}.PQ=sin80∘sin30∘​=2sin80∘1​. So α=12sin⁡80∘.\alpha=\frac{1}{2\sin80^\circ}.α=2sin80∘1​. Also, PR=sin⁡70∘sin⁡80∘.PR=\frac{\sin70^\circ}{\sin80^\circ}.PR=sin80∘sin70∘​.

  1. Find angles in triangle PRSPRSPRS

Given: ∠PRS=40∘.\angle PRS=40^\circ.∠PRS=40∘. At PPP, ∠QPS=∠QPR+∠RPS=80∘+θ.\angle QPS=\angle QPR+\angle RPS=80^\circ+\theta.∠QPS=∠QPR+∠RPS=80∘+θ. But in triangle PQSPQSPQS, since ∠PQS=15∘\angle PQS=15^\circ∠PQS=15∘, we will not need this directly.

In triangle PRSPRSPRS, the third angle is ∠PSR=180∘−40∘−θ=140∘−θ.\angle PSR=180^\circ-40^\circ-\theta=140^\circ-\theta.∠PSR=180∘−40∘−θ=140∘−θ.

  1. Use area relation via triangle PQSPQSPQS

Now observe that [PQS]=[PQR]+[PRS].[PQS]=[PQR]+[PRS].[PQS]=[PQR]+[PRS]. Also,

=\frac12\alpha\beta\sin(80^\circ+\theta).$$ So $$\frac12\alpha\beta\sin(80^\circ+\theta)=\frac12\alpha\sin70^\circ+[PRS].$$ This route is possible but cumbersome. A cleaner way is to use triangle $PQS$ directly. In triangle $PQS$: - $\angle PQS=15^\circ$ - $\angle QPS=80^\circ+\theta$ - hence $$\angle PSQ=180^\circ-(15^\circ+80^\circ+\theta)=85^\circ-\theta.$$ Using sine rule in triangle $PQS$, $$\frac{PS}{\sin15^\circ}=\frac{PQ}{\sin(85^\circ-\theta)}.$$ Thus $$\beta=\alpha\cdot \frac{\sin15^\circ}{\sin(85^\circ-\theta)}.$$ This still contains $\theta$. So let us instead determine $\theta$ from triangle $PRS$ using $PR$ and $RS=1$? Actually, $RS$ is not given. We need another relation. 6. **Find $RS$ from triangle $QRS$ and connect with triangle $PRS$** In triangle $QRS$, we know only $QR=1$ and angle at $R$ is $70^\circ$, but no side/angle at $Q,S$ directly. So this is not enough. Let us place coordinates. 7. **Coordinate setup** Take $$Q=(0,0),\quad R=(1,0).$$ Since $\angle PQR=70^\circ$, point $P$ lies on the ray making $70^\circ$ with positive $x$-axis from $Q$: $$P=(\alpha\cos70^\circ,\alpha\sin70^\circ).$$ At $R$, $\angle QRS=70^\circ$ means $RS$ makes angle $110^\circ$ with positive $x$-axis, so let $$S=(1+t\cos110^\circ,\ t\sin110^\circ),$$ for some $t=RS>0$. Now use $\angle PRS=40^\circ$. Vector $$\overrightarrow{RP}=(\alpha\cos70^\circ-1,\alpha\sin70^\circ).$$ But from triangle $PQR$, we already found $\angle PRQ=30^\circ$, so the ray $RP$ makes angle $150^\circ$ with positive $x$-axis. Hence line $RP$ is fixed, and since line $RS$ has direction $110^\circ$, angle between them is indeed $40^\circ$. Thus $P$ lies on the ray from $R$ at angle $150^\circ$ with length $PR=\dfrac{\sin70^\circ}{\sin80^\circ}$. So $$P=\left(1+PR\cos150^\circ,\ PR\sin150^\circ\right).$$ Now line $QS$ makes angle $15^\circ$ with line $QP$, and since $QP$ has direction $70^\circ$, line $QS$ has direction $55^\circ$ from positive $x$-axis. Therefore $S$ is intersection of: - ray from $Q$ at angle $55^\circ$ - ray from $R$ at angle $110^\circ$ So $\triangle QRS$ has: $$\angle Q=55^\circ,\quad \angle R=70^\circ,\quad \angle S=55^\circ.$$ Hence it is isosceles with $$RS=QR=1.$$ So $t=1$. 8. **Now compute $[PRS]$** Since $$PR=\frac{\sin70^\circ}{\sin80^\circ},\qquad RS=1,\qquad \angle PRS=40^\circ,$$ area of triangle $PRS$ is $$[PRS]=\frac12\,PR\cdot RS\sin40^\circ =\frac12\cdot \frac{\sin70^\circ}{\sin80^\circ}\sin40^\circ.$$ But also $$[PRS]=\frac12\beta\cdot PR\sin\theta.$$ Therefore $$\frac12\beta\,PR\sin\theta= rac12\,PR\sin40^\circ,$$ so $$\beta\sin\theta=\sin40^\circ.$$ 9. **Compute the target expression** We need $$4\alpha\beta\sin\theta=4\alpha(\beta\sin\theta)=4\alpha\sin40^\circ.$$ Using $$\alpha=\frac{1}{2\sin80^\circ},$$ we get $$4\alpha\beta\sin\theta=4\cdot \frac{1}{2\sin80^\circ}\cdot \sin40^\circ =\frac{2\sin40^\circ}{\sin80^\circ}.$$ Now use $$\sin80^\circ=2\sin40^\circ\cos40^\circ,$$ so $$\frac{2\sin40^\circ}{\sin80^\circ}=\frac{2\sin40^\circ}{2\sin40^\circ\cos40^\circ}=\sec40^\circ.$$ Thus $$4\alpha\beta\sin\theta=\sec40^\circ.$$ 10. **Locate $\sec40^\circ$ in the given intervals** Numerically, $$\cos40^\circ\approx 0.7660,$$ so $$\sec40^\circ\approx 1.305.$$ Now check options: - **A:** $(0,\sqrt2)=(0,1.414\ldots)$ → contains $1.305$ ✅ - **B:** $(1,2)$ → contains $1.305$ ✅ - **C:** $(\sqrt2,3)=(1.414\ldots,3)$ → does not contain $1.305$ ❌ - **D:** $(2\sqrt2,3\sqrt2)$ → far larger, does not contain ❌ So the correct options are $$\boxed{A, B}.$$
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