Question

Let $PQR$ be a triangle of area $△$ 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

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

Answer 1

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


We have,
$a=2,b=\frac{7}{2}$ and $c=\frac{5}{2}$
$\therefore s=4$
$△=\sqrt{4\cdot 2\cdot \frac{1}{2}\cdot \frac{3}{2}}=\sqrt{6}$
Now,
$\Rightarrow \frac{2\sin P-\sin 2P}{2\sin P+\sin 2P}$
$=\frac{2\sin P-2\sin P\cos P}{2\sin P+2\sin P\cos P}$
$=\frac{1-\cos P}{1+\cos P}$
$={\tan }^2\frac{P}{2}$
$={\left(\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\right)}^2$
$=\frac{\frac{1}{2}\cdot \frac{3}{2}}{4\cdot 2}=\frac{3}{32}={\left(\frac{3}{4△}\right)}^2$