Question

a line intersects two parallel lines, forming four acute angles and four obtuse angles. the measure of one of the acute angles is (7x - 410)°. the sum of the measures of one of the acute angles and three of the obtuse angles is (-14x + w)°. what is the value of w?

Explanation:

Step1: Recall angle - relationships

When a line intersects two parallel lines, corresponding angles are equal, and the sum of an acute - angle and an obtuse - angle is 180°. Also, the four acute angles are equal and the four obtuse angles are equal. Let the measure of one acute angle be \(A=(7x - 410)^{\circ}\) and the measure of one obtuse angle be \(O=(-14x + w)^{\circ}\). Since \(A + O=180^{\circ}\), we have the equation \((7x-410)+(-14x + w)=180\).

Step2: Simplify the equation

Combine like - terms: \(7x-14x+w-410 = 180\). This simplifies to \(-7x+w-410 = 180\).

Step3: Isolate \(w\)

Add \(7x\) and 410 to both sides of the equation. We get \(w=7x + 590\).
However, we also know that the measure of an acute angle \(7x-410>0\) (since it's an acute angle, its measure is between 0 and 90). Solving \(7x-410>0\) gives \(7x>410\), so \(x > \frac{410}{7}\approx58.57\). Also, since the angle is acute, \(7x-410 < 90\), which gives \(7x<500\), so \(x<\frac{500}{7}\approx71.43\).
Since the angles are formed by a transversal intersecting parallel lines, we can also use the fact that the sum of the measures of one of the acute angles and three of the obtuse angles is a multiple of 180°. But a simpler approach is to assume the acute and obtuse angles are supplementary.
Let's assume we want to find \(x\) first from the acute - angle formula. Since \(7x-410\) is an acute angle, we know that \(7x-410>0\). Let's assume the acute angle is non - zero.
If we consider the fact that the sum of an acute and an obtuse angle formed by a transversal intersecting parallel lines is 180°.
We have \(7x-410+(-14x + w)=180\).
We can also note that when two parallel lines are intersected by a transversal, the acute angles are equal and the obtuse angles are equal.
Let's assume the angle relationship based on supplementary angles:
\[

$$\begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=180 + 410\\ -7x+w&=590\\ w&=7x + 590 \end{align*}$$

\]
If we assume the acute angle \(7x-410\) is a valid acute - angle measure. Let's say the acute angle \(7x - 410\) is non - negative. If we assume the acute angle is at its maximum value (close to 90°), \(7x-410 = 90\), then \(7x=500\), \(x=\frac{500}{7}\approx71.43\).
If we assume the acute angle is at its minimum non - zero value, \(7x-410 = 1\), then \(7x=411\), \(x=\frac{411}{7}\approx58.71\).
Let's go back to the equation \(-7x + w=590\).
We know that the sum of an acute and an obtuse angle formed by a transversal of parallel lines is 180°.
Since the acute angle is \(7x-410\) and the obtuse angle is \(-14x + w\), we have:
\[

$$\begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590 \end{align*}$$

\]
If we assume the acute angle \(7x - 410\) is a valid angle measure. Let's find \(x\) from the acute - angle property. Since \(7x-410>0\), \(x>\frac{410}{7}\approx58.57\).
Let's assume the acute angle \(7x - 410 = 10\) (a valid acute - angle value), then \(7x=420\), \(x = 60\).
Substitute \(x = 60\) into the equation \(w=7x + 590\).
\[

$$\begin{align*} w&=7\times60+590\\ &=420+590\\ &=1010 \end{align*}$$

\]
Let's re - check using the angle relationships. The acute angle is \(7x-410=7\times60 - 410=420 - 410 = 10^{\circ}\), and the obtuse angle is \(-14\times60+1010=-840 + 1010 = 170^{\circ}\), and \(10 + 170=180^{\circ}\)

Answer:

\(w = 1010\)