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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Mathematics
Chapter
Properties of Triangles

JEE Advanced 2012 · Paper 2 · Mathematics · Properties of Triangles

Let P Q R be a triangle of area Δ with a=2, b=(7)/(2) and c=(5)/(2)...

Let PQRP Q R be a triangle of area Δ\Delta with a=2,b=72a=2, b=\frac{7}{2} and c=52;c=\frac{5}{2} ; where a,ba, b, and cc are the lengths of the sides of the triangle opposite to the angles at P,QP, Q and RR respectively. Then 2sin⁡P−sin⁡2P2sin⁡P+sin⁡2P\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P} equals.

Options

  1. A34Δ\frac{3}{4 \Delta}
  2. B454Δ\frac{45}{4 \Delta}
  3. C(34Δ)2\left(\frac{3}{4 \Delta}\right)^{2}
  4. D(454Λ)2\left(\frac{45}{4 \Lambda}\right)^{2}

Correct answer

C. (34Δ)2\left(\frac{3}{4 \Delta}\right)^{2}

Step-by-step solution

2sin⁡P−sin⁡2P2sin⁡P+sin⁡2P=2sin⁡P−2sin⁡Pcos⁡P2sin⁡P+2sin⁡Pcos⁡P=1−cos⁡P1−sin⁡P\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}=\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}=\frac{1-\cos P}{1-\sin P} =2sin⁡2P22cos⁡2P2=tan⁡2P2=(s−b)(s−c)s(s−a)=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)} where s=a+b+c2s=\frac{a+b+c}{2} =(s−b)2(s−c)2s(s−a)(s−b)(s−c)=(a+c−b)2(a+b−c)216.Δ2=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}} =(2+52−72)2(2+72−52)216Δ2=1×916Δ2=(34Δ)2=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}
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