If ABC is a triangle of area Delta with a = 2 | vedclass

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(B) Step $1$: Simplify the expression $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A} = \frac{2 \sin A - 2 \sin A \cos A}{2 \sin A + 2 \sin A \cos A} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Step $1$: Simplify the expression $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A} = \frac{2 \sin A - 2 \sin A \cos A}{2 \sin A + 2 \sin A \cos A} = \frac{2 \sin A (1 - \cos A)}{2 \sin A (1 + \cos A)} = \frac{1 - \cos A}{1 + \cos A} = 	an^2(\frac{A}{2})$.Step $2$: Use the half-angle formula $	an^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{s(s-a)}$, where $s = \frac{a+b+c}{2}$.Step $3$: Calculate $s = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$.Step $4$: Substitute values: $	an^2(\frac{A}{2}) = \frac{(4 - 3.5)(4 - 2.5)}{4(4 - 2)} = \frac{0.5 	imes 1.5}{4 	imes 2} = \frac{0.75}{8} = \frac{3}{32}$.Step $5$: Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{4(4-2)(4-3.5)(4-2.5)} = \sqrt{4 	imes 2 	imes 0.5 	imes 1.5} = \sqrt{6}$.Step $6$: Check options: $(\frac{3}{4\Delta})^2 = \frac{9}{16 	imes 6} = \frac{9}{96} = \frac{3}{32}$. Thus, the answer is $(\frac{3}{4\Delta})^2$.

If $ABC$ is a triangle of area $\Delta$ with $a = 2$, $b = \frac{7}{2}$, $c = \frac{5}{2}$, where $a, b, c$ are the lengths of the sides of the triangle opposite to angles $A, B$ and $C$ respectively, then $\frac{2 \sin A - \sin 2A}{2 \sin A + \sin 2A}$ is equal to...

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