Question

If $PQR$be a triangle of the 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 $\frac{(2\sin P-\sin 2P)}{(2\sin P+\sin 2P)}$ is equal to

• A. $\frac{3}{4}∆$
• B. $\frac{45}{4}∆$
• C. ${\left(\frac{3}{4}∆\right)}^2$
• D. ${\left(\frac{45}{4}∆\right)}^5$

Answer 1

The correct option is C

${\left(\frac{3}{4}∆\right)}^2$

Explanation for the correct option
Step 1: Find a semi-perimeter for given $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}$

If $PQR$ be a triangle of the area ∆ and $a=2,b=\frac{7}{2},$$c=\frac{5}{2}$ (where $a,b,c$ the length of the sides of the triangle)

$\left(\because s=\frac{a+b+c}{2}\right)$

Where $s$ is the semi-perimeter of a triangle

$$\begin{gathered} \therefore s=\frac{\left(2+{\displaystyle \frac{7}{2}}+{\displaystyle \frac{5}{2}}\right)}{2}=\frac{{\displaystyle \left(\frac{4+7+5}{2}\right)}}{2} \\ \Rightarrow s=\left(\frac{16}{4}\right) \\ \Rightarrow s=4 \end{gathered}$$

Step 2: Find the value of $\frac{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{P}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{P}\mathbf{)}}{\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{P}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{P}\mathbf{)}}$

Solve above expression

$\therefore \frac{(2\sin P-\sin 2P)}{(2\sin P+\sin 2P)}$

$=\frac{(2\sin P-2\sin P\cos P)}{(2\sin P+2\sin P\cos P)}$ $(\because \sin 2P=2\sin P\cos P)$

$=\frac{2\sin P(1-\cos P)}{2\sin P(1+\cos P)}$

$=\frac{\left(1-\cos P\right)}{\left(1+\cos P\right)}$

Now solve the above equation from the formula of $\cos 2A$

$(\because \cos 2A={\cos }^{2}A-{\sin }^{2}A=2{\cos }^{2}A-1=1-2{\sin }^{2}A)$

But here $(\cos A={\cos }^2\frac{A}{2}-{\sin }^2\frac{A}{2}=2{\cos }^2\frac{A}{2}-1=1-2{\sin }^2\frac{A}{2})$

$\therefore \frac{2{\sin }^2{\displaystyle \left(\frac{P}{2}\right)}}{2{\cos }^2{\displaystyle \left(\frac{P}{2}\right)}}$ $\left(\because \frac{\sin {\displaystyle A}}{\cos A}=\tan A\right)$

$={\tan }^2\left(\frac{P}{2}\right)$

$\left\{\because tan\left(\frac{P}{2}\right)=\sqrt{\frac{\left(\right(s–b\left)\right(s–c)}{s(s–a)}}\right\}$ (From Properties of triangle formulae)

$\therefore \frac{(s–b)(s–c)}{s(s–a)}$

Multiply numerator and denominator by (s – b)(s – c)

$=\frac{{(s–b)}^2{(s–c)}^2}{s(s–a)(s-b)(s-c)}$

Step 3: Put the values of $\mathit{s}\mathbf{,}\mathit{a}\mathbf{,}\mathit{b}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{c}$ in the above equation.

$\frac{{(s–b)}^2{(s–c)}^2}{s(s–a)(s-b)(s-c)}$

$=\frac{{\left(4–{\displaystyle \frac{7}{2}}\right)}^2{\left(4–{\displaystyle \frac{5}{2}}\right)}^2}{{△}^2}$ $\left\{\because △=\sqrt{s(s-a)(s-b)(s-c)}\right\}$

$$\begin{gathered} =\frac{{\left({\displaystyle \frac{1}{2}}\right)}^2{\left({\displaystyle \frac{3}{2}}\right)}^2}{{△}^2} \\ ={\left(\frac{3}{4△}\right)}^2 \end{gathered}$$

Hence option (C) is the correct answer.