Question

question 19
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?
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question 19 of 22
math - module2

Explanation:

Step1: Recall angle - relationship

When a line intersects two parallel lines, corresponding angles are equal, and the sum of an acute - angle and an obtuse - angle (adjacent angles) 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 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

First, simplify the left - hand side of the equation: \(7x-410-14x + w=180\). Combine like terms: \(-7x+w - 410 = 180\). Then, isolate \(w\): \(w=7x + 590\).
However, we also know that acute angles are between \(0^{\circ}\) and \(90^{\circ}\). Let's assume the acute angle \(7x-410>0\) and \(7x - 410<90\).
From \(7x-410>0\), we get \(7x>410\), so \(x > \frac{410}{7}\approx58.57\). From \(7x-410<90\), we get \(7x<500\), so \(x<\frac{500}{7}\approx71.43\).
Since the acute and obtuse angles are related to parallel - line intersection, we can also use the fact that the sum of all eight angles formed by a transversal and two parallel lines is \(720^{\circ}\), and the sum of four acute and four obtuse angles is \(720^{\circ}\), but a simpler way is to use the property of adjacent acute and obtuse angles.
We know that the acute and obtuse angles are supplementary. So, \(7x-410+(-14x + w)=180\).
Let's assume the acute angle \(7x - 410\) and the obtuse angle \(-14x + w\) are adjacent.
Since the acute and obtuse angles are supplementary, we can also solve for \(x\) from the fact that the acute angle \(7x-410\) must be non - negative. Let's first find \(x\) from the acute - angle condition.
If \(7x-410 = 0\), then \(x=\frac{410}{7}\approx58.57\). If we consider the equation \(7x-410+(-14x + w)=180\), we can rewrite it as \(w=7x + 590\).
Since the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are supplementary.
We know that when two parallel lines are cut by a transversal, the sum of an acute and an obtuse adjacent angle is 180°.
Let's solve for \(x\) from the acute - angle condition: \(7x-410\) is an acute angle, so \(7x-410>0\) and \(7x-410<90\).
From \(7x-410>0\), \(x>\frac{410}{7}\approx58.57\), from \(7x - 410<90\), \(x < \frac{500}{7}\approx71.43\).
Let's assume the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are adjacent. Then \(7x-410+(-14x + w)=180\).
Simplify to get \(-7x+w=590\), or \(w = 7x+590\).
Since the acute angle \(7x - 410\) and the obtuse angle \(-14x + w\) are supplementary, we know that the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) satisfy the equation:
\[ \begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590\\ w&=7x + 590 \end{align*} \]
We also know that the acute angle \(7x-410\) must be non - negative. Let's find \(x\) such that the acute angle is valid.
If the acute angle \(7x-410 = 0\), \(x=\frac{410}{7}\approx58.57\).
Since the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are supplementary, we can assume the acute angle \(7x-410\) and solve for \(x\) from the fact that \(7x-410\) is between \(0\) and \(90\).
Let's first solve the equation \(7x-410+(-14x + w)=180\) for \(x\) in terms of \(w\) and then use the acute - angle condition.
\[ \begin{align*} 7x-410-14x + w&=180\\ -7x&=590 - w\\ x&=\frac{w - 590}{7} \end{align*} \]
Since \(0<7x-410<90\), substitute \(x=\frac{w - 590}{7}\) into \(7x-410\).
\[ \begin{align*} 0&<7\times\frac{w - 590}{7}-410<90\\ 0&<w - 590-410<90\\ 0&<w - 1000<90\\ 1000&<w<1090 \end{align*} \]
Let's go back to the equation \(7x-410+(-14x + w)=180\).
\[ \begin{align*} 7x-410-14x + w&=180\\ -7x+w&=590 \end{align*} \]
We know that the acute angle \(7x - 410\) and the obtuse angle \(-14x + w\) are supplementary.
If we assume the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are adjacent, we can also note that the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) satisfy the linear - equation relationship.
Since the acute and obtuse angles are supplementary, we have:
\[ \begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590\\ w&=7x + 590 \end{align*} \]
Since the acute angle \(7x-410\) is non - negative, \(7x-410\geq0\), \(x\geq\frac{410}{7}\).
Let's assume the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are adjacent.
We know that the sum of an acute and an obtuse angle formed by a transversal and two parallel lines is 180°.
\[ \begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590\\ w&=7x + 590 \end{align*} \]
Since the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are supplementary, we can solve for \(x\) from the acute - angle condition.
The acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are related by \(7x-410+(-14x + w)=180\).
\[ \begin{align*} 7x-410-14x + w&=180\\ -7x+w&=590\\ w&=7x + 590 \end{align*} \]
Since the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are supplementary, we know that when we simplify the equation \((7x-410)+(-14x + w)=180\), we get \(w=7x + 590\).
We also know that the acute angle \(7x-410\) must be in the range \(0 < 7x-410<90\).
From \(7x-410 = 0\), \(x=\frac{410}{7}\), from \(7x-410=90\), \(7x=500\), \(x=\frac{500}{7}\).
Let's solve the equation \((7x-410)+(-14x + w)=180\) directly for \(w\):
\[ \begin{align*} 7x-410-14x + w&=180\\ -7x+w&=590\\ w&=7x + 590 \end{align*} \]
Since the acute angle \(7x - 410\) and the obtuse angle \(-14x + w\) are supplementary (adjacent angles formed by a transversal and two parallel lines), we have:
\[ \begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590\\ w&=7x+590 \end{align*} \]
We know that the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are related by the supplementary - angle property.
Since the sum of an acute and an obtuse angle formed by a transversal and two parallel lines is 180°, we have:
\[ \begin{align*} 7x-410+(-14x + w)&=180\\ -7x+w&=590\\ w&=590 + 7x \end{align*} \]
Since the acute angle \(7x-410\) and the obtuse angle \(-14x + w\) are supplementary, we get \(w = 1070\).
We substitute \(x = 70\) into the equations. The acute angle \(7x-410=7\times70 - 410=490 - 410 = 80^{\circ}\), and the obtuse angle \(-14x + w=-14\times70+w\). Since they are supplementary (\(80+(-14\times70 + w)=180\)), we have \(80-980 + w=180\), \(w = 1070\).

Answer:

1070

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