Geometry

Well, it's been a while since I've answered a question on this website, and what better way to celebrate with a geometry question :)

So, we don't have a right triangle, so we're gonna have to settle with Law of Cosines. 

Also, note that we have 

$(cos PMQ) = (- cos PMR )$

Now, let's apply Law of Cosines and add subsitute using equation above. 

$PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)$

Simplify

$5^2 = 5.5^2 + PM^2 + 2(5.5 * PM)cos(PMR) \\ 8^2 = 5.5^2 + PM^2 - 2(5,5 * PM) cos (PMR)$

Now, we're stuck...but we can add these two equations to keep progressing. 

$5^2 + 8^2 = 2 * 5.5^2 + 2PM^2 \\ 89 = 60.5 + 2PM^2 \\ 28.5 / 2 = PM^2\\ 14.25 = PM^2 \\ \sqrt {14.25} = PM ≈ 3.77$

And we're done!

Also, side note, this is gonna be my 1000 answer on this website which is a weird flex but whatever. 

I might be gone for another while, but hopefully be back in the summer. 

~NTS