Question

1. in the figure, $overrightarrow{tp}$ and $overrightarrow{ts}$ are opposite rays. $overrightarrow{tq}$ bisects $angle rtp$. if $mangle ptq = 12x + 4$ and $mangle rtq = 15x - 5$, find $mangle rtp$.
2. in the figure, $overrightarrow{cj}$ and $overrightarrow{cr}$ are opposite rays. $overrightarrow{cp}$ bisects $angle ncw$ and $overrightarrow{cl}$ bisects $angle jcn$. if $mangle jcl = 6x + 3$ and $mangle lcn = 8x - 13$, find $mangle jcl$.
3. if $mangle pcw = 4b - 6$ and $mangle wcr = 3b + 2$, find $mangle wcr$.
4. if $mangle ncp = 5k - 8$ and $mangle ncw = 8k + 2$, find $mangle pcw$.
5. if $mangle jcn = 8r - 2$ and $mangle lcn = 3r + 15$, find $mangle jcl$.

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{TQ}$ bisects $\angle RTP$, then $m\angle PTQ=m\angle RTQ$. So, set up the equation $12x + 4=15x-5$.

Step2: Solve the equation for $x$

Subtract $12x$ from both sides: $4 = 3x-5$. Then add 5 to both sides: $9 = 3x$. Divide both sides by 3, we get $x = 3$.

Step3: Find $m\angle RTP$

Since $m\angle RTP=m\angle PTQ + m\angle RTQ$ and $m\angle PTQ=m\angle RTQ$, $m\angle RTP=2m\angle PTQ$. Substitute $x = 3$ into $m\angle PTQ=12x + 4$, we have $m\angle PTQ=12\times3+4=36 + 4=40$. So $m\angle RTP=2\times40 = 80$.

Answer:

$m\angle RTP = 80$