Chapter 10.5, Problem 18PSB

a.

To determine

To find $\angle R$

Answer to Problem 18PSB

  $m\angle R={57.5}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle R=\frac{m\stackrel{⌢}{PQ}}{2}\\ m\angle R=\frac{{115}^{\circ }}{2}\\ m\angle R={57.5}^{\circ }\end{array}$

b.

To determine

To find $\angle S$

Answer to Problem 18PSB

  $m\angle S={57.5}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle S=\frac{m\stackrel{⌢}{PQ}}{2}\\ m\angle S=\frac{{115}^{\circ }}{2}\\ m\angle S={57.5}^{\circ }\end{array}$

c.

To determine

To find $\stackrel{⌢}{SR}$

Answer to Problem 18PSB

  $m\stackrel{⌢}{SR}={72}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle RPS=\frac{m\stackrel{⌢}{SR}}{2}\\ {36}^{\circ }=\frac{m\stackrel{⌢}{SR}}{2}\\ m\stackrel{⌢}{SR}={72}^{\circ }\end{array}$

d.

To determine

To find $\stackrel{⌢}{QR}$

Answer to Problem 18PSB

  $m\stackrel{⌢}{QR}={108}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\stackrel{⌢}{SR}+m\stackrel{⌢}{QR}=180\\ m\stackrel{⌢}{QR}=180-{72}^{\circ }\\ m\stackrel{⌢}{QR}={108}^{\circ }\end{array}$

e.

To determine

To find $\angle QPR$

Answer to Problem 18PSB

  $m\angle QPR={54}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle QPR=\frac{m\stackrel{⌢}{QR}}{2}\\ m\angle QPR=\frac{{108}^{\circ }}{2}\\ m\angle QPR={54}^{\circ }\end{array}$

f.

To determine

To find $\angle QPS$

Answer to Problem 18PSB

  $m\angle QPS={90}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle QPS=\frac{m\stackrel{⌢}{QRS}}{2}\\ m\angle QPS=\frac{{180}^{\circ }}{2}\\ m\angle QPS={90}^{\circ }\end{array}$

g.

To determine

To find $\angle QTP$

Answer to Problem 18PSB

  $m\angle QTP={93.5}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle QTP=\frac{m\stackrel{⌢}{PQ}+m\stackrel{⌢}{SR}}{2}\\ m\angle QTP=\frac{{115}^{\circ }+{72}^{\circ }}{2}\\ m\angle QTP={93.5}^{\circ }\end{array}$

h.

To determine

To find $\angle PQV$

Answer to Problem 18PSB

  $m\angle PQV={57.5}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\angle PQV=\frac{m\stackrel{⌢}{PQ}}{2}\\ m\angle PQV=\frac{{115}^{\circ }}{2}\\ m\angle PQV={57.5}^{\circ }\end{array}$

i.

To determine

To find $m\stackrel{⌢}{PRQ}$

Answer to Problem 18PSB

  $m\stackrel{⌢}{PRQ}={245}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\stackrel{⌢}{PRQ}=360-m\stackrel{⌢}{PR}\\ m\stackrel{⌢}{PRQ}=360-115\\ m\stackrel{⌢}{PRQ}={245}^{\circ }\end{array}$

j.

To determine

To find $m\stackrel{⌢}{RSP}$

Answer to Problem 18PSB

  $m\stackrel{⌢}{RSP}={137}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}m\stackrel{⌢}{RSP}=360-m\stackrel{⌢}{PQ}-m\stackrel{⌢}{QR}\\ m\stackrel{⌢}{RSP}=360-115-108\\ m\stackrel{⌢}{RSP}={137}^{\circ }\end{array}$

k.

To determine

To find $\angle QVS$

Answer to Problem 18PSB

  $m\angle QVS={90}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}SQ\perp VQ\\ m\angle QVS={90}^{\circ }\end{array}$

l.

To determine

To find $\angle QOP$

Answer to Problem 18PSB

  $\angle QOP={115}^{\circ }$ .

Explanation of Solution

Given:

Given: $\overleftrightarrow{VQ}$ is tangent to $\odot O$ at Q.

  $\overline{QS}$ is a diameter of $\odot O$ .

  $\widehat{PQ}=115°$ , $\angle RPS=36°$

  

Calculation:

  $\begin{array}{l}\angle QOP=m\stackrel{⌢}{PQ}\\ \angle QOP={115}^{\circ }\end{array}$