Chapter 1.5, Problem 20PSB

a.

To determine

To-Determine: The $m\angle QPT$ .

Answer to Problem 20PSB

  $m\angle QPT=94°$

Explanation of Solution

Given:

  

  $\begin{array}{l}\angle RPS=23°50\text{'}\\ \overrightarrow{PR}\text{and }\overrightarrow{PS}\text{trisect }\angle QPT\end{array}$

Concept Used:

When angle is trisected each angle has equal measure.

Calculation:

Here, we have

  $\begin{array}{l}\angle RPS=23°50\text{'}\\ \overrightarrow{PR}\text{and }\overrightarrow{PS}\text{trisect }\angle QPT\\ \Rightarrow \angle QPR=23°50\text{'},\angle RPS=23°50\text{'},\angle SPT=23°50\text{'}\\ \because \angle QPT=\angle QPR+\angle RPS+\angle SPT\\ \Rightarrow \angle QPT=23°50\text{'}+23°50\text{'}+23°50\text{'}\\ \Rightarrow \angle QPT=69°150\text{'}\\ \Rightarrow \angle QPT=69°+\left(\frac{150}{6}\right)°\\ \Rightarrow \angle QPT=69°+25°\\ \Rightarrow \angle QPT=94°\end{array}$

Conclusion:

  $\angle QPT=94°$

b.

To determine

To-Determine: The $m\angle QPS$ .

Answer to Problem 20PSB

  $m\angle QPS=80°32\text{'}20"$

Explanation of Solution

Given:

  

  $\begin{array}{l}m\angle QPT=120°48\text{'}30"\\ \overrightarrow{PR}\text{and }\overrightarrow{PS}\text{trisect }\angle QPT\end{array}$

Concept Used:

When angle is trisected each angle has equal measure.

Calculation:

Here, we have

  $\begin{array}{l}m\angle QPT=120°48\text{'}30"\\ \overrightarrow{PR}\text{and }\overrightarrow{PS}\text{trisect }\angle QPT\\ \Rightarrow \angle QPR=\angle RPS=\angle SPT=\frac{1}{3}\times 120°48\text{'}30"\\ \Rightarrow \angle QPR=\angle RPS=\angle SPT=40°16\text{'}10"\\ \because \angle QPS=\angle QPR+\angle RPS\\ \Rightarrow \angle QPS=40°16\text{'}10"+40°16\text{'}10"\\ \Rightarrow \angle QPS=80°32\text{'}20"\end{array}$

Conclusion:

  $\angle QPS=80°32\text{'}20"$