Chapter 10.6, Problem 17PSB

a.

To determine

Find the value for the given angle using diagram.

Answer to Problem 17PSB

The value of the angle $\angle P={29}^{°}$ .

Explanation of Solution

Given:

The given diagram is

  

The given data are $\angle S={88}^{°}$ , $ArcQT={104}^{°},ArcST={94}^{°}$ . The given angle is $\angle P$ .

Calculation:

Use the inscribed angle theorem $\to$ the measure of an inscribed angle is half the measure of its intercepted arc.

  $\begin{array}{l}m\angle S=\frac{1}{2}marcTQR\\ \Rightarrow 2m\angle S=marcTQR\\ \Rightarrow 2m\angle S=marcQT+marcRQ\\ \Rightarrow 2m\angle S-marcQT=marcQT+marcRQ-marcQT\\ \Rightarrow 2m\angle S-marcQT=marcRQ\\ \Rightarrow marcRQ=2\cdot {88}^{°}-{104}^{°}\\ \Rightarrow marcRQ={176}^{°}-{104}^{°}\\ \Rightarrow marcRQ={72}^{°}\end{array}$

Now,

  $\begin{array}{l}mArcQT+mArcRQ+mArcSR+mArcST={360}^{°}\\ \Rightarrow mArcSR={360}^{°}-mArcQT-mArcRQ-mArcST\\ \Rightarrow mArcSR={360}^{°}-{104}^{°}-{72}^{°}-{94}^{°}\\ \Rightarrow mArcSR={90}^{°}\end{array}$

Angle formed by tangent and secant $=\frac{1}{2}$

  $=$ difference of intercepted arcs

  $\begin{array}{l}m\angle P=\frac{1}{2}\left(ArcSQ-ArcQT\right)\\ =\frac{1}{2}\left(ArcSR+ArcRQ-ArcQT\right)\\ =\frac{1}{2}\left({90}^{°}+{72}^{°}-{104}^{°}\right)\\ =\frac{{58}^{°}}{2}\\ ={29}^{°}\end{array}$

Hence the value of the angle $\angle P={29}^{°}$ .

b.

To determine

Find the value for the given angle using diagram.

Answer to Problem 17PSB

The value of the angle $\angle STQ={81}^{°}$ .

Explanation of Solution

Given:

The given diagram is

  

The given data are $\angle S={88}^{°}$ , $ArcQT={104}^{°},ArcST={94}^{°}$ . The given angle is $\angle STQ$ .

Calculation:

Use the inscribed angle theorem $\to$ the measure of an inscribed angle is half the measure of its intercepted arc.

  $\begin{array}{l}m\angle S=\frac{1}{2}marcTQR\\ \Rightarrow 2m\angle S=marcTQR\\ \Rightarrow 2m\angle S=marcQT+marcRQ\\ \Rightarrow 2m\angle S-marcQT=marcQT+marcRQ-marcQT\\ \Rightarrow 2m\angle S-marcQT=marcRQ\\ \Rightarrow marcRQ=2\cdot {88}^{°}-{104}^{°}\\ \Rightarrow marcRQ={176}^{°}-{104}^{°}\\ \Rightarrow marcRQ={72}^{°}\end{array}$

Now,

  $\begin{array}{l}mArcQT+mArcRQ+mArcSR+mArcST={360}^{°}\\ \Rightarrow mArcSR={360}^{°}-mArcQT-mArcRQ-mArcST\\ \Rightarrow mArcSR={360}^{°}-{104}^{°}-{72}^{°}-{94}^{°}\\ \Rightarrow mArcSR={90}^{°}\end{array}$

The measure of an inscribed angle is half the measure of its intercepted arc.

  $\begin{array}{l}m\angle STQ=\frac{1}{2}marcSRQ\\ =\frac{1}{2}\left(marcSR+marcRQ\right)\\ =\frac{1}{2}\left({90}^{°}+{72}^{°}\right)\\ =\frac{{162}^{°}}{2}\\ ={81}^{°}\end{array}$

Hence the value of the angle $\angle STQ={81}^{°}$ .