Let PQR be a triangle of area Delta with a=2, | vedclass

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(A) Given $a=2, b=\frac{7}{2}, c=\frac{5}{2}$.Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.The expression is $\fra

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$\left(\frac{3}{4 \Delta}\right)^2$

Vedclass · Answer

(A) Given $a=2, b=\frac{7}{2}, c=\frac{5}{2}$.Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.The expression is $\frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P} = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P} = \frac{2 \sin P(1 - \cos P)}{2 \sin P(1 + \cos P)} = \frac{1 - \cos P}{1 + \cos P}$.Using half-angle formulas,$\frac{1 - \cos P}{1 + \cos P} = \frac{2 \sin^2(P/2)}{2 \cos^2(P/2)} = 	an^2(P/2)$.We know $	an^2(P/2) = \frac{(s-b)(s-c)}{s(s-a)}$.Substituting the values: $s-a = 4-2 = 2$,$s-b = 4-3.5 = 0.5 = 1/2$,$s-c = 4-2.5 = 1.5 = 3/2$.$	an^2(P/2) = \frac{(1/2)(3/2)}{4(2)} = \frac{3/4}{8} = \frac{3}{32}$.Also,$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{4 	imes 2 	imes 0.5 	imes 1.5} = \sqrt{6}$.Thus,$\left(\frac{3}{4 \Delta}ight)^2 = \frac{9}{16 \Delta^2} = \frac{9}{16 	imes 6} = \frac{9}{96} = \frac{3}{32}$.Therefore,the expression equals $\left(\frac{3}{4 \Delta}ight)^2$.

Let $PQR$ 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 $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2P}$ equals

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