Discussion Forum

Let PQR  be a triangle of area $\triangle$ with a=2,b=$\frac{7}{2}$ and c=$\frac{5}{2}$ where a,b and c  the length of the sides of the triangle opposite to the angle s  at P, Q and R respectively.Then $\frac{2 \sin P-\sin 2P}{2 \sin P+\sin 2P}$ equal to 

Answer:

Explanation:

 Concept involved if $\triangle ABC$ has sides a, b, c then, 

 $\tan (A/2)=\sqrt{\frac{(s-b)(s-c)}{a(s-a)}}$

 where, $s=\frac{a+b+c}{2}$

 Sol ,$s= \frac{2+\frac{7}{2}+\frac{5}{2}}{2}=4$

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

 $= \frac{2 \sin^{2}(P/2)}{2 \cos^{2}(P/2)}=\tan^{2}(P/2)$

 $\Rightarrow$      $\frac{(s-b)(s-c)}{s(s-a)} \times \frac{(s-b)(s-c)}{(s-b)(s-c)}$

 $\Rightarrow$     $\frac{((s-b)(s-c)^{2})}{\triangle^{2}}= \frac{\left(4+\frac{7}{2}\right)^{2}\left(4-\frac{5}{2}\right)^{2}}{\triangle^{2}}$

 =$(\frac{3}{4 \triangle})^{2}$