Question

a line intersects two parallel lines, forming four acute angles and four obtuse angles. the measure of one of the acute angles is (9x - 650). the sum of the measure of one of the acute angles and three of the obtuse angles is (-18x + w). what is the value of w?

Explanation:

Step1: Recall angle properties of parallel lines

When a transversal intersects two parallel lines, acute and obtuse angles are supplementary (sum to \(180^\circ\)), and all acute angles are equal, all obtuse angles are equal. There are 4 acute and 4 obtuse angles. So sum of all acute angles: \(4\times(9x - 650)\), sum of three obtuse angles: \(-18x + w\), and the fourth obtuse angle is \(180 - (9x - 650)\) (since acute + obtuse = 180). But also, sum of all obtuse angles should be \(4\times(180 - (9x - 650))\). However, a simpler approach: each acute angle and each obtuse angle are supplementary, so all acute angles are equal, all obtuse angles are equal. Let acute angle be \(A = 9x - 650\), obtuse angle be \(B = 180 - A\). There are 4 acute and 4 obtuse angles. The sum of one acute and three obtuse angles: \(A + 3B=-18x + w\). Substitute \(B = 180 - A\): \(A + 3(180 - A)=-18x + w\). Simplify left side: \(A + 540 - 3A = 540 - 2A\). Substitute \(A = 9x - 650\): \(540 - 2(9x - 650)=-18x + w\). Expand: \(540 - 18x + 1300=-18x + w\). Combine like terms: \(1840 - 18x=-18x + w\).

Step2: Solve for \(w\)

Subtract \(-18x\) from both sides: \(1840 = w\).

Answer:

\(1840\)