Chapter 5.3, Problem 26PSC

a.

To determine

To Find: How many pairs of angles are formed?

Answer to Problem 26PSC

14 pairs of angles are formed, when two parallel lines are intersected by a transversal excluding linear pair of angles.

Explanation of Solution

Given:

  

  $a\parallel b$ and eight angles are formed here.

Concept Used:

When two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

Calculation:

Here, we have

  

  $a\parallel b$ and eight angles are formed here.

Since, we know that when two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

  $\begin{array}{l}\text{Since, we know that when two parallel lines are intersected by a transversal following angles formed:}\\ \text{Pairs of corresponding angles: }\angle 1\text{and }\angle 5,\angle 2\text{and }\angle 6,\angle 3\text{and }\angle 8,\angle 4\text{and }\angle 7\\ \text{Pairs of alternate interior angles:}\angle 4\text{and }\angle 5,\angle 3\text{and }\angle 6\\ \text{Pairs of alternate exterior angles:}\angle 1\text{and }\angle 7,\angle 2\text{and }\angle 8\\ \text{Pairs of supplementary angles:}\angle 3\text{and }\angle 5,\angle 4\text{and }\angle 6\\ \text{Pairs of vertically opposite angles: }\angle 1\text{and }\angle 4,\angle 2\text{and }\angle 3,\angle 5\text{and }\angle 7,\angle 6\text{and }\angle 8\end{array}$

Thus, there are 14 pairs of angles formed, when two parallel lines intersected by a transversal.

b.

To determine

To Find: The probability of choosing pairs of alternate interior or pairs of exterior angles of pairs of corresponding angles.

Answer to Problem 26PSC

Probability of choosing pairs of alternate interior or pairs of exterior angles of pairs of corresponding angles $=\frac{4}{7}$ ..

Explanation of Solution

Given:

  

  $a\parallel b$ and eight angles are formed here.

Concept Used:

When two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

Calculation:

Here, we have

  

  $a\parallel b$ and eight angles are formed here.

Since, we know that when two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

  $\begin{array}{l}\text{Since, we know that when two parallel lines are intersected by a transversal following angles formed:}\\ \text{Pairs of corresponding angles: }\angle 1\text{and }\angle 5,\angle 2\text{and }\angle 6,\angle 3\text{and }\angle 8,\angle 4\text{and }\angle 7\\ \text{Pairs of alternate interior angles:}\angle 4\text{and }\angle 5,\angle 3\text{and }\angle 6\\ \text{Pairs of alternate exterior angles:}\angle 1\text{and }\angle 7,\angle 2\text{and }\angle 8\\ \text{Pairs of supplementary angles:}\angle 3\text{and }\angle 5,\angle 4\text{and }\angle 6\\ \text{Pairs of vertically opposite angles: }\angle 1\text{and }\angle 4,\angle 2\text{and }\angle 3,\angle 5\text{and }\angle 7,\angle 6\text{and }\angle 8\end{array}$ Thus, there are 14 pairs of angles formed, when two parallel lines intersected by a transversal.

  $\begin{array}{l}\because \text{Pairs of alternate interior angles:}\angle 4\text{and }\angle 5,\angle 3\text{and }\angle 6\\ \therefore \text{Probability of choosing pairs of alternate interior angles}=\frac{2}{14}\\ \text{Pairs of corresponding angles: }\angle 1\text{and }\angle 5,\angle 2\text{and }\angle 6,\angle 3\text{and }\angle 8,\angle 4\text{and }\angle 7\\ \therefore \text{Probability of choosing pairs of corresponding angles}=\frac{4}{14}\\ \because \text{Pairs of alternate exterior angles:}\angle 1\text{and }\angle 7,\angle 2\text{and }\angle 8\\ \therefore \text{Probability of choosing pairs of alternate exterior angles}=\frac{2}{14}\\ \therefore \text{Probability of choosing pairs of alternate interior angles }\\ \text{ or pairs of corresponding angles}\text{or pairs of alternate exterior angles}\\ =\frac{2}{14}\text{+ }\frac{4}{14}+\frac{2}{14}=\frac{8}{14}=\frac{4}{7}\end{array}$

Hence, the probability of choosing pairs of alternate interior or pairs of exterior angles of pairs of corresponding angles $=\frac{4}{7}$ .

c.

To determine

To Find: The probability of choosing pairs of supplementary angles.

Answer to Problem 26PSC

  $\text{The probability of choosing pairs of supplementary angles}=\frac{1}{7}$

Explanation of Solution

Given:

  

  $a\parallel b$ and eight angles are formed here.

Concept Used:

When two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

Calculation:

Here, we have

  

  $a\parallel b$ and eight angles are formed here.

Since, we know that when two parallel lines are intersected by a transversal following angles formed:

• Pair of corresponding angles
• Pair of alternate interior angles
• Pair of alternate exterior angles
• Pair of supplementary angles
• Pair of vertically opposite angles

  $\begin{array}{l}\text{Since, we know that when two parallel lines are intersected by a transversal following angles formed:}\\ \text{Pairs of corresponding angles: }\angle 1\text{and }\angle 5,\angle 2\text{and }\angle 6,\angle 3\text{and }\angle 8,\angle 4\text{and }\angle 7\\ \text{Pairs of alternate interior angles:}\angle 4\text{and }\angle 5,\angle 3\text{and }\angle 6\\ \text{Pairs of alternate exterior angles:}\angle 1\text{and }\angle 7,\angle 2\text{and }\angle 8\\ \text{Pairs of supplementary angles:}\angle 3\text{and }\angle 5,\angle 4\text{and }\angle 6\\ \text{Pairs of vertically opposite angles: }\angle 1\text{and }\angle 4,\angle 2\text{and }\angle 3,\angle 5\text{and }\angle 7,\angle 6\text{and }\angle 8\end{array}$ Thus, there are 14 pairs of angles formed, when two parallel lines intersected by a transversal.

  $\begin{array}{l}\because \text{Pairs of supplementary angles:}\angle 3\text{and }\angle 5,\angle 4\text{and }\angle 6\\ \therefore \text{Probability of choosing pairs of supplementary angles}=\frac{2}{14}=\frac{1}{7}\end{array}$

  $\text{Hence, the probability of choosing pairs of supplementary angles}=\frac{1}{7}$