Question

vertical and adjacent angles · practice
example 1
refer to the figure.

1. name two adjacent angles
2. name two vertical angles
3. find ( mangle suv ).

find the value of each variable.

1. figure: two intersecting lines, one angle ( 120^circ ), another ( (2x - 10)^circ )
2. figure: two intersecting lines (one vertical), angles ( (2x)^circ ) and ( (4x + 108)^circ )

Explanation:

Problem 4:

Step1: Identify vertical angles

Vertical angles are equal. The angle of \(120^\circ\) and the angle \((2x - 10)^\circ\) are vertical angles? Wait, no, adjacent to \(120^\circ\) is a supplementary angle. Wait, actually, the angle supplementary to \(120^\circ\) and \((2x - 10)^\circ\) – no, wait, vertical angles: when two lines intersect, vertical angles are equal. Wait, the \(120^\circ\) angle and the angle opposite to \((2x - 10)^\circ\)? Wait, no, let's see: the angle adjacent to \(120^\circ\) is \(180 - 120 = 60^\circ\)? Wait, no, maybe I made a mistake. Wait, the angle \((2x - 10)^\circ\) and the angle that is vertical to the angle supplementary to \(120^\circ\)? Wait, no, let's start over.

When two lines intersect, adjacent angles are supplementary (sum to \(180^\circ\)), and vertical angles are equal. Wait, the \(120^\circ\) angle and the angle \((2x - 10)^\circ\) – are they vertical angles? Wait, no, maybe the \(120^\circ\) angle and the angle adjacent to \((2x - 10)^\circ\) are supplementary. Wait, no, the correct approach: the angle of \(120^\circ\) and the angle \((2x - 10)^\circ\) – wait, no, vertical angles are equal. Wait, maybe the \(120^\circ\) angle and the angle opposite to it (vertical angle) is equal, but here we have \((2x - 10)^\circ\). Wait, no, perhaps the angle adjacent to \(120^\circ\) is \(180 - 120 = 60^\circ\), and that angle is equal to \((2x - 10)^\circ\) because they are vertical angles. Ah, that makes sense. So:

Step1: Set up the equation

The angle supplementary to \(120^\circ\) is \(180 - 120 = 60^\circ\). This angle is vertical to \((2x - 10)^\circ\), so they are equal. Thus:
\(2x - 10 = 60\)

Step2: Solve for \(x\)

Add 10 to both sides:
\(2x = 60 + 10\)
\(2x = 70\)

Divide both sides by 2:
\(x = \frac{70}{2}\)
\(x = 35\)

Wait, but let's check. If \(x = 35\), then \(2x - 10 = 70 - 10 = 60\), which is supplementary to \(120^\circ\) (60 + 120 = 180), so that works.

Problem 5:

Step1: Identify the relationship

The angles \((2x)^\circ\) and \((4x + 108)^\circ\) are adjacent and form a linear pair? Wait, no, there's a vertical line, so maybe they are supplementary? Wait, no, the angle \((2x)^\circ\) and \((4x + 108)^\circ\) – wait, looking at the diagram, one is a horizontal line, one is a vertical line? Wait, no, the diagram shows a vertical line and a horizontal line? Wait, no, the angles \((2x)^\circ\) and \((4x + 108)^\circ\) are adjacent and form a linear pair? Wait, no, maybe they are supplementary? Wait, no, if there's a vertical line, then the angle between the horizontal and vertical is \(90^\circ\), but here we have \((2x)^\circ\) and \((4x + 108)^\circ\) – wait, maybe they are supplementary? Wait, no, let's think again.

Wait, the two angles \((2x)^\circ\) and \((4x + 108)^\circ\) – are they adjacent and form a linear pair? Wait, no, maybe they are supplementary? Wait, no, if they are on a straight line (linear pair), their sum is \(180^\circ\). Wait, but also, there's a vertical line, so maybe one of them is \(90^\circ\)? Wait, no, the diagram shows a vertical line and a horizontal line intersecting, so the angle between them is \(90^\circ\), but here we have \((2x)^\circ\) and \((4x + 108)^\circ\) – wait, maybe they are supplementary? Wait, no, let's check the equation.

Wait, the angles \((2x)^\circ\) and \((4x + 108)^\circ\) – are they vertical angles? No, vertical angles are equal. Wait, maybe they are supplementary? Wait, no, if they are adjacent and form a linear pair, their sum is \(180^\circ\). Wait, but let's set up the equation:

Step1: Set up the equation

Assuming they are supplementary (sum to \(180^\circ\)):
\(2x + (4x + 108) = 180\)

Step2: Solve for \(x\)

Combine like terms:
\(6x + 108 = 180\)

Subtract 108 from both sides:
\(6x = 180 - 108\)
\(6x = 72\)

Divide both sides by 6:
\(x = \frac{72}{6}\)
\(x = 12\)

Wait, but let's check. If \(x = 12\), then \(2x = 24\), and \(4x + 108 = 48 + 108 = 156\). Then \(24 + 156 = 180\), which is correct for a linear pair. So that works.

Answer:

s:

1. \(x = 35\)
2. \(x = 12\)