Question

Let $PQR$ be a triangle of area $\Delta$ with $a=2,b=\frac{7}{2},c=\frac{5}{2},$ 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 the value of $\frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}$ is :

• A. $\frac{3}{4\Delta }$
• B. $\frac{45}{4\Delta }$
• C. ${\left(\frac{3}{4\Delta }\right)}^2$
• D. ${\left(\frac{45}{4\Delta }\right)}^2$

Answer 1

The correct option is C ${\left(\frac{3}{4\Delta }\right)}^2$

Given: $a=2,b=\frac{7}{2},c=\frac{5}{2},$
$2s=a+b+c=2+\frac{7}{2}+\frac{5}{2}=8$
$s=4$
$\Rightarrow \frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}$ $=\frac{2\sin P(1-\cos P)}{2\sin P(1+\cos P)}$
$=\frac{2{\sin }^2\frac{P}{2}}{2{\cos }^2\frac{P}{2}}$
$={\tan }^2\frac{P}{2}$
$={\left(\frac{(s-b)(s-c)}{\Delta }\right)}^2$
$=\left(\frac{(4-3.5)(4-2.5)}{\Delta }\right)$
$={\left(\frac{0.75}{\Delta }\right)}^2$
$={\left(\frac{3}{4\Delta }\right)}^2$