problem is for testing skills

Here's my solution

. Hexagon named ABMCDE side lengths of BM =5, MC =3, CD =6, DE =7, EA =X,AB=Y. We need to find X and Y sum of all interior angles of hexagon is 720 degrees.
Then each interior angles would be 120 degrees. we know that the exterior angle +interior angle is 180 degrees . If we extend the lines you could see that how the figure looks like . The triangles are EQUILATERAL THOSE WHICH ARE DRAWN IN CORNER. If we look at the biggest triangle PQR is also an equilateral triangle. NOTE: PB=Y So,here the algebra comes that each side of a is 16. 16=X+Y+7 16=Y+5+3 hence,x=1 y=8

Use vector algebra. Let $a$ and $b$ be the unknown lengths. Then

$5+3\cos 60\degree+6\cos 120\degree+7\cos 180\degree+a\cos 240\degree+b\cos 300\degree=0$

$\implies a-b=-7$

$3\sin 60\degree+6\sin 120\degree+7\sin 180\degree+a\sin 240\degree+b\sin 300\degree=0$

$\implies a+b=9$

So, $a=\dfrac 12(-7+9)=\boxed 1$ ,

$b=\dfrac 12(9+7)=\boxed 8$ .