In a triangle PQR ,P is the largest angle and cos P = `1/3` . further the triangle toches the side 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)

Options

A. 16,18
B. 18,22
C. 22,24
D. 16,20

Explanation

Let PN = n, QL = n + 2 and RM = n + 4. Then,
PM =PN =n, QN =QL =n+ 2 and RL =RM =n + 4
`therefore` PQ = n+(n+2) = 2n+2,QR=(n+2)+(n+4)=2n+6 and RP =(n + 4) +n=2n + 4
`thereforecosP=(PQ^(2)+PR^(2)-QR^(2))/(2PQ.PR)`
`rArrcosP=((2n+2)^(2)+(2n+4)^(2)-(2n+6)^(2))/(2(2n+2)(2n+4))=(4n^(2)-16)/(8(n+1)(n+2))`
But`,cosP=1/3` [Given]
`therefore(4n^(2)-16)/(8(n+1)(n+2))=1/3`
`rArr(n-2)/(2(n+1))=1/3`
`rArr3n-6=2n+2`
`rArr3n-6=2n+2`
`rArrn=8`
`thereforePq=18,QR=22` and RP=20