−→ bisects <NLM, LM = 12, NO = 3 and LN = 8. What is the value of x? Drawing is not to scale. Please show ALL work in order to receive full credit for the problem
We are given that $NO=3$, $LN=8$, and $LM = 12$. We are asked to find the value of $x$.
First, draw a diagram and label the given lengths as follows:
[asy]
pair L,M,N;
L = (0,0);
M = (12,0);
N = (0,10);
draw(L--M--N--cycle);
draw(L--N,dashed);
label("$L$",L,SW);
label("$M$",M,SE);
label("$N$",N,N);
label("$8$",(L+N)/2,W);
label("$12$",(L+M)/2,S);
[/asy]
Since line $LN$ bisects angle $\angle LNM$, it cuts it into two congruent angles. Therefore, we have $\angle LNM = \angle MLN$.
Next, we can use the Law of Cosines in triangle $LMN$ to relate the sides and angles. We have:
\begin{align*}
MN^2 &= LM^2 + LN^2 - 2(LM)(LN)\cos(\angle LNM).
\end{align*}From the given values, we can substitute in $12$ for $LM$, $8$ for $LN$, and $10$ for $MN$. We can simplify the equation to solve for $\cos(\angle LNM) = \cos(x)$ as follows:
\begin{align*}
100 &= 144 + 64 - 2(12)(8)\cos(x)\\
100 &= 208 - 192\cos(x)\\
-108 &= -192\cos(x)\\
\cos(x) &= \frac{108}{192}\\
\cos(x) &= \frac{9}{16}.
\end{align*}Since $\cos(x) = \cos(\angle LNM)$, we can conclude that $\angle LNM = \angle MLN \approx 49.06^\circ$.
Finally, the value of $x$ is equal to $\angle MNO = 180^\circ - \angle LNM = 180^\circ - 49.06^\circ = \boxed{130.94 ^\circ}$.