Question

#12: m∠gkh = ____ °
(13x - 1)°
(9x + 3)°
your answer
#13: m∠mnq = ____ °
(2x - 1)°
(5x - 40)°

Explanation:

Step1: Identify angle - relationship for #12

Assume $\angle{GKH}$ and $\angle{JKG}$ are supplementary (a common situation if they form a linear - pair). So, $(13x - 1)+(9x + 3)=180$.
\[

$$\begin{align*} 13x-1 + 9x+3&=180\\ 22x+2&=180\\ 22x&=180 - 2\\ 22x&=178\\ x&=\frac{178}{22}=\frac{89}{11}\approx8.09 \end{align*}$$

\]
Then, $m\angle{GKH}=(9x + 3)^{\circ}$. Substitute $x = \frac{89}{11}$ into the expression:
\[

$$\begin{align*} m\angle{GKH}&=9\times\frac{89}{11}+3\\ &=\frac{801}{11}+3\\ &=\frac{801 + 33}{11}\\ &=\frac{834}{11}\approx75.82^{\circ} \end{align*}$$

\]

Step2: Identify angle - relationship for #13

Assume $\angle{MNQ}$ and $\angle{PNR}$ are vertical angles (since they are opposite each other when two lines intersect). So, $2x-1 = 5x - 40$.
\[

$$\begin{align*} 2x-1&=5x - 40\\ 40 - 1&=5x-2x\\ 39&=3x\\ x& = 13 \end{align*}$$

\]
Then, $m\angle{MNQ}=(2x - 1)^{\circ}$. Substitute $x = 13$ into the expression: $m\angle{MNQ}=2\times13-1=26 - 1=25^{\circ}$.

Answer:

#12: $\frac{834}{11}\approx75.82^{\circ}$
#13: $25^{\circ}$