Chapter 4.4, Problem 22WE

(a)

To determine

Find the measures of $\angle 2,\angle 7,\angle 5,\text{ and }\angle \text{6}$ .Decide whether $\overline{PQ}\parallel \overline{SR}$ .

Answer to Problem 22WE

  $m\angle 2=m\angle 5=m\angle \text{6=40}°\text{ and m}\angle \text{7=100}°$

Explanation of Solution

Given:

  $\overline{PO}\cong \overline{QO};\overline{RO}\cong \overline{SO}$

  $m\angle 1=40°$

Calculation:

Represent the given situation:

  

Since, $\overline{PO}\cong \overline{QO}$ , so the triangle POQ is isosceles .

  $m\angle 1=m\angle 2=40°$

Now, $m\angle 7=180°-40°-40°=100°$

Since $\angle 7\text{ and }\angle \text{O}$ are vertically opposite angles, so they are congruent.

Hence, $m\angle 7=m\angle O=100°$

Since $\overline{RO}\cong \overline{SO}$ , so the triangle SOR is isosceles.

  $m\angle 5=m\angle 6=\frac{180°-100°}{2}=40°$

Now , since $m\angle 1=m\angle 6\Rightarrow \angle 1\cong \angle 6$ and are alternate interior angles where PR is the transversal to the line segments $\overline{PQ}\text{ and }\overline{SR}$ , so , by the Converse of Alternate Interior Angle Theorem , $\overline{PQ}\parallel \overline{SR}$ .

So, $m\angle 2=m\angle 5=m\angle \text{6=40}°\text{ and m}\angle \text{7=100}°$

(b)

To determine

Find the measures of $\angle 2,\angle 7,\angle 5,\text{ and }\angle \text{6}$ .Decide whether $\overline{PQ}\parallel \overline{SR}$ .

Answer to Problem 22WE

  $m\angle 2=m\angle 5=m\angle \text{6=k}°\text{ and m}\angle \text{7=180}°-2k°$

Explanation of Solution

Given:

  $\overline{PO}\cong \overline{QO};\overline{RO}\cong \overline{SO}$

  $m\angle 1=k$

Calculation:

Represent the given situation:

  

Since, $\overline{PO}\cong \overline{QO}$ , so the triangle POQ is isosceles .

  $m\angle 1=m\angle 2=k°$

Now, $m\angle 7=180°-k°-k°=180°-2k°$

Since $\angle 7\text{ and }\angle \text{O}$ are vertically opposite angles, so they are congruent.

Hence, $m\angle 7=m\angle O=180°-2k°$

Since $\overline{RO}\cong \overline{SO}$ , so the triangle SOR is isosceles.

  $m\angle 5=m\angle 6=\frac{180°-(180°-2k°)}{2}=k°$

Now , since $m\angle 1=m\angle 6\Rightarrow \angle 1\cong \angle 6$ and are alternate interior angles where PR is the transversal to the line segments $\overline{PQ}\text{ and }\overline{SR}$ , so , by the Converse of Alternate Interior Angle Theorem , $\overline{PQ}\parallel \overline{SR}$ .

So, $m\angle 2=m\angle 5=m\angle \text{6=k}°\text{ and m}\angle \text{7=180}°-2k°$