In a triangle PQR,P is the largest angle and | vedclass

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(B) Let the lengths of the tangents from the vertices be $x, y, z$. Given $PN=x, QL=y, RM=z$ are consecutive even integers. Let $x=2k, y=2k+2, z=2k+4$

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$(B, D)$

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(B) Let the lengths of the tangents from the vertices be $x, y, z$. Given $PN=x, QL=y, RM=z$ are consecutive even integers. Let $x=2k, y=2k+2, z=2k+4$.Since $P$ is the largest angle,the side opposite to $P$ $(QR = y+z)$ must be the largest side.$QR = (2k+2) + (2k+4) = 4k+6$$PQ = x+y = 2k + 2k+2 = 4k+2$$PR = x+z = 2k + 2k+4 = 4k+4$Using the Law of Cosines: $\cos P = \frac{PQ^2 + PR^2 - QR^2}{2(PQ)(PR)} = \frac{1}{3}$$\frac{(4k+2)^2 + (4k+4)^2 - (4k+6)^2}{2(4k+2)(4k+4)} = \frac{1}{3}$$\frac{16k^2+16k+4 + 16k^2+32k+16 - (16k^2+48k+36)}{2(16k^2+24k+8)} = \frac{1}{3}$$\frac{16k^2}{32k^2+48k+16} = \frac{1}{3} \Rightarrow \frac{k^2}{2k^2+3k+1} = \frac{1}{3}$$3k^2 = 2k^2+3k+1 \Rightarrow k^2-3k-1 = 0$. This gives no integer solution for $k$.Re-evaluating the consecutive even integers as $2n, 2n+2, 2n+4$ where $PN=2n, QL=2n+2, RM=2n+4$:$PQ = 4n+2, PR = 4n+4, QR = 4n+6$. $\cos P = \frac{(4n+2)^2+(4n+4)^2-(4n+6)^2}{2(4n+2)(4n+4)} = \frac{1}{3}$$\frac{16n^2+16n+4+16n^2+32n+16-16n^2-48n-36}{2(16n^2+24n+8)} = \frac{1}{3}$ $\Rightarrow \frac{16n^2-16}{32n^2+48n+16} = \frac{1}{3}$$\frac{n^2-1}{2n^2+3n+1} = \frac{1}{3}$ $\Rightarrow 3n^2-3 = 2n^2+3n+1$ $\Rightarrow n^2-3n-4 = 0$$(n-4)(n+1) = 0 \Rightarrow n=4$.Sides are $PQ = 18, PR = 20, QR = 22$. The possible lengths are $18$ and $22$.

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