Question

in the figure, $overrightarrow{cj}$ and $overrightarrow{cr}$ are opposite rays. $overrightarrow{cp}$ bisects $angle ncw$ and $overrightarrow{cl}$ bisects $angle jcn$. if $mangle pcw = 7g$ and $mangle wcr = 5g + 14$, find $mangle wcr$.

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{CP}$ bisects $\angle NCW$, then $m\angle PCW=m\angle PCN = 7g$.

Step2: Consider linear - pair of angles

$\angle JCR$ is a straight - angle, so $m\angle JCL + m\angle LCN+m\angle NCP + m\angle PCW+m\angle WCR=180^{\circ}$. Also, since $\overrightarrow{CL}$ bisects $\angle JCN$, we can ignore it for the relationship between $\angle PCW$ and $\angle WCR$. We know that $\angle PCW$ and $\angle WCR$ are part of a linear - pair (or the sum of angles around point $C$). And since $\angle PCW$ and $\angle WCR$ are adjacent angles and $\angle JCR$ is a straight - angle, we have $m\angle PCW + m\angle WCR=90^{\circ}$ (assuming $\angle JCR$ is a straight - angle and we are dealing with right - angle related angle bisections in a half - of the straight - angle situation). Substituting the given expressions: $7g+(5g + 14)=90$.

Step3: Solve the equation for $g$

Combine like terms: $7g+5g+14 = 90$. So, $12g+14 = 90$. Subtract 14 from both sides: $12g=90 - 14=76$. Then $g=\frac{76}{12}=\frac{19}{3}$.

Step4: Find $m\angle WCR$

Substitute $g = \frac{19}{3}$ into the expression for $m\angle WCR$. $m\angle WCR=5g + 14$. So, $m\angle WCR=5\times\frac{19}{3}+14=\frac{95}{3}+14=\frac{95 + 42}{3}=\frac{137}{3}\approx45.67^{\circ}$.

Answer:

$\frac{137}{3}$