JEE Advanced · Mathematics · 17. Properties of Triangles
Let \(P Q R\) be a triangle of area \(\Delta\) with \(a=2, b=\frac{7}{2}\) and \(c=\frac{5}{2} ;\) where \(a, b\), and \(c\) are the lengths of the sides of the triangle opposite to the angles at \(P, Q\) and \(R\) respectively. Then \(\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}\) equals.
- A \(\frac{3}{4 \Delta}\)
- B \(\frac{45}{4 \Delta}\)
- C \(\left(\frac{3}{4 \Delta}\right)^{2}\)
- D \(\left(\frac{45}{4 \Lambda}\right)^{2}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{3}{4 \Delta}\right)^{2}\)
Step-by-step Solution
Detailed explanation
\(\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}=\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}=\frac{1-\cos P}{1-\sin P}\)
\(=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}\) where \(s=\frac{a+b+c}{2}\)
\(=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}\)
\(=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}\)
\(=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}\) where \(s=\frac{a+b+c}{2}\)
\(=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}\)
\(=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}\)
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