Question

#1.) in the figure below, $overline{ad}$ is the angle bisector of $angle cab$ and $mangle eaf = 180^{circ}$. if $mangle cab=(x^{2}-9)^{circ}$ and $mangle dab=(4x)^{circ}$, what is $mangle dac$? #2.) in the figure, $mangle jkm = 180^{circ}$, and $overline{kn}$ bisects $angle jkl$. if $mangle jkn=(8x - 13)^{circ}$, and $mangle nkl=(6x + 11)^{circ}$, find $angle nkm$. *#3.) in the diagram below $overline{cf}$ bisects $angle dcb$. if $mangle fcd=(7x)^{circ}$ and $mangle fcb=(6x^{2}-10)^{circ}$. determine the $mangle dcf$.

Explanation:

Step1: Use angle - bisector property for #1

Since $\overline{AD}$ is the angle - bisector of $\angle CAB$, then $m\angle DAC=m\angle DAB$ and $m\angle CAB = 2m\angle DAB$. So we set up the equation $x^{2}-9 = 2\times(4x)$.
\[x^{2}-9 = 8x\]
\[x^{2}-8x - 9=0\]
Factor the quadratic equation: \((x - 9)(x+1)=0\). So \(x = 9\) or \(x=-1\). Since angle measures are non - negative, we take \(x = 9\). Then \(m\angle DAC=m\angle DAB=(4x)^{\circ}=(4\times9)^{\circ}=36^{\circ}\).

Step2: Use angle - bisector property for #2

Since $\overline{KN}$ bisects $\angle JKL$, then $m\angle JKN=m\angle NKL$. Set up the equation \(8x - 13=6x + 11\).
\[8x-6x=11 + 13\]
\[2x=24\]
\[x = 12\]. Then \(m\angle NKM=m\angle JKN=(8x - 13)^{\circ}=(8\times12-13)^{\circ}=(96 - 13)^{\circ}=83^{\circ}\).

Step3: Use angle - bisector property for #3

Since $\overline{CF}$ bisects $\angle DCB$, then $m\angle FCD=m\angle FCB$. Set up the equation \(7x=6x^{2}-10\).
\[6x^{2}-7x - 10=0\]
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 6\), \(b=-7\), \(c=-10\)).
\[x=\frac{7\pm\sqrt{(-7)^{2}-4\times6\times(-10)}}{2\times6}=\frac{7\pm\sqrt{49 + 240}}{12}=\frac{7\pm\sqrt{289}}{12}=\frac{7\pm17}{12}\]
We get \(x=\frac{7 + 17}{12}=\frac{24}{12}=2\) or \(x=\frac{7-17}{12}=-\frac{5}{6}\). Since angle measures are non - negative, we take \(x = 2\). Then \(m\angle DCF=(7x)^{\circ}=(7\times2)^{\circ}=14^{\circ}\).

Answer:

1. \(m\angle DAC = 36^{\circ}\)
2. \(m\angle NKM = 83^{\circ}\)
3. \(m\angle DCF = 14^{\circ}\)