Discussion Forum

 In a $\triangle$ PQR, P is the largest angle and  $\cos p=\frac{1}{3}$ . Further in circle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of Pn, QL, and RM are consecutive even integers. Then  possible length(s) of the side(s)  of the triangle is (are)

Answer:

Explanation:

Concept involved

 When ever cosine of angle and sides are given or to find out , we should always use cosine law

 i.e,    $\cos A= \frac{b^{2}+c^{2}-a^{2}}{2bc}$

     $\cos B= \frac{a^{2}+c^{2}-b^{2}}{2ac}$

$\cos C= \frac{a^{2}+b^{2}-c^{2}}{2ba}$

$\therefore$   $\cos P= \frac{b^{2}+c^{2}-a^{2}}{2bc}$

 $\Rightarrow \frac{1}{3}=\frac{(2n+4)^{2}+(2n+2)^{2}-(2n+6)^{2}}{2(2n+4)(2n+2)}$

    ($\because$    cos P= $\frac{1}{3}$)

 $\Rightarrow \frac{4n^{2}-16}{8(n+1)(n+2)}=\frac{1}{3}$

  $\Rightarrow \frac{n^{2}-4}{2(n+1)(n+2)}=\frac{1}{3}$

   $\Rightarrow \frac{(n-2)}{2(n+1)}=\frac{1}{3}$

$\Rightarrow$     3n-6=2n+2 $\Rightarrow$ n=8

$\therefore$   sides are 2n+2, 2n+4,2n+6

$\Rightarrow$  18,20,22