for the angle PLN, the hypotenuse PL = 15, the cosine of it is 4/5,
what about the length of NL? well, NL would be the "adjacent" side
of the angle PLN, so, recall your SOH CAH TOA
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)
which of those fellows, give us only
the angle
adjacent side
and
hypotenuse?
well, is Ms Cosine, so let's bother her
\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad
\begin{cases}
cos(PLN)=\frac{4}{5}\\
hypotenuse=PL=15
\end{cases}
\\\\
thus
\\\\
cos(PLN)=\cfrac{NL}{PL}\implies \cfrac{4}{5}=\cfrac{NL}{PL}\implies \cfrac{4\cdot PL}{5}=NL
\\\\
\cfrac{4\cdot 15}{5}=NL\)
now, that we know what NL is, what is ML?
notice, the angle MLK is really the same shared angle PLN
so the cosine of that is the same 4/5
so let us use cosine again, now for the bigger triangle though
\(\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad
\begin{cases}
adjacent=KL=KN+NL\\
cos(MLK)=\frac{4}{5}
\end{cases}
\\\\\\
cos(MLK)=\cfrac{KN+NL}{ML}\implies \cfrac{4}{5}=\cfrac{4+NL}{ML}\implies
ML=\cfrac{5(4+NL)}{4}\)
so.. now we know what the sides of KL and ML are,
but what about MK? well
notice, is a right-triangle,
you have the hypotenuse ML,
you also have the adjacent side, KL,
to find the opposite side of MK, simply use the pythagorean theorem
\(\bf c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}\)
