Given that  
BA
  and  
BD
  are opposite rays, m∠CBA=12n+4, and m∠DBC=14n−6, find m∠ABC.

Given that \( \overrightarrow{BA} \) and \( \overrightarrow{BD} \) are opposite rays, this means that the angles \( \angle CBA \) and \( \angle DBC \) are adjacent angles on a straight line. The sum of the measures of these angles must therefore be 180°, because they form a linear pair.

Given:
- \( m\angle CBA = 12n + 4 \)
- \( m\angle DBC = 14n - 6 \)

Since \( \angle CBA \) and \( \angle DBC \) form a straight line, we can set up the following equation:

\[
(12n + 4) + (14n - 6) = 180
\]

Now, let's solve for \( n \):

1. Combine like terms:

\[
26n - 2 = 180
\]

2. Add 2 to both sides:

\[
26n = 182
\]

3. Divide both sides by 26:

\[
n = 7
\]

Now that we know \( n = 7 \), we can substitute it back into the expressions for \( m\angle CBA \) and \( m\angle DBC \):

- \( m\angle CBA = 12(7) + 4 = 84 + 4 = 88^\circ \)
- \( m\angle DBC = 14(7) - 6 = 98 - 6 = 92^\circ \)

Thus, the measure of \( \angle ABC \), which is the entire straight line, is:

\[
m\angle ABC = m\angle CBA + m\angle DBC = 88^\circ + 92^\circ = 180^\circ
\]

This confirms that the angles form a linear pair, and \( m\angle ABC \) is \( 180^\circ \).

Would you like more details or have any questions? Here are 5 related questions to expand on this topic:

1. What are opposite rays, and how do they relate to straight angles?
2. How do linear pairs of angles help in solving angle problems?
3. What is the difference between adjacent and supplementary angles?
4. How would the problem change if the angles were complementary instead of forming a straight line?
5. Can you explain the relationship between linear pairs and vertical angles?

**Tip:** Always remember that opposite rays create a straight angle, meaning the sum of the angles must equal 180°.

Explore how to solve an angle problem involving opposite rays and linear pairs. Understand the relationship between adjacent angles on a straight line, and how to calculate angle measures using algebraic expressions. Suitable for students in Grades 9-10.

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