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Difficulty: MediumAngles, Parallel Lines, and Polygons

Line ABAB is parallel to line CDCD. A transversal line EFEF intersects line ABAB at point PP and line CDCD at point QQ. If ∠APQ=(4x−10)∘\angle APQ = (4x - 10)^\circ and ∠PQD=(2x+30)∘\angle PQD = (2x + 30)^\circ are alternate interior angles, what is the measure of ∠BPQ\angle BPQ?

  1. 110∘110^\circAnswer
  2. B
    70∘70^\circ
  3. C
    20∘20^\circ
  4. D
    100∘100^\circ

Answer

110∘110^\circ
Since alternate interior angles are equal, 4x−10=2x+304x - 10 = 2x + 30, yielding x=20x = 20. Thus, ∠APQ=70∘\angle APQ = 70^\circ. Because ∠APQ\angle APQ and ∠BPQ\angle BPQ are adjacent angles on straight line ABAB, their sum is 180∘180^\circ, giving ∠BPQ=110∘\angle BPQ = 110^\circ.

Step-by-Step Solution

1
Set up the equation for alternate interior angles.
4x−10=2x+304x - 10 = 2x + 30
Alternate interior angles formed by a transversal cutting parallel lines are equal.
2
Solve the linear equation for xx.
2x=40  ⟹  x=202x = 40 \implies x = 20
Subtract 2x2x and add 1010 to both sides.
3
Calculate the measure of ∠APQ\angle APQ.
\angle APQ = 4(20) - 10 = 70^\circ$
Substitute x=20x = 20 into the expression (4x−10)∘(4x - 10)^\circ.
4
Determine ∠BPQ\angle BPQ using the straight line angle property.
\angle BPQ = 180^\circ - 70^\circ = 110^\circ$
Angles ∠APQ\angle APQ and ∠BPQ\angle BPQ form a linear pair on straight line ABAB, summing to 180∘180^\circ.

Key Concept

Alternate interior angles of parallel lines and angles on a straight line
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Topics

Angles, Parallel Lines, and Polygons

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