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Difficulty: Very hardLines and Angles

In the geometric plane, line kk is parallel to line mm (k∥mk \parallel m). Transversal line tt intersects line kk at point AA and line mm at point BB. Point CC lies on line kk to the left of AA, such that interior acute angle ∠CAB=(3x−10)∘\angle CAB = (3x - 10)^\circ. Ray ADAD bisects ∠CAB\angle CAB. Ray BFBF is drawn into the region between lines kk and mm making an angle ∠ABF=(x+35)∘\angle ABF = (x + 35)^\circ with transversal segment ABAB. Ray ADAD and ray BFBF intersect at point PP inside the parallel region. Line PBPB is extended past PP to intersect line kk at point QQ. If ray ADAD is perpendicular to ray BFBF, what is the measure of the obtuse angle formed at the intersection of line QBQB and line kk?

  1. 121∘121^\circAnswer
  2. B
    59∘59^\circ
  3. C
    118∘118^\circ
  4. D
    128∘128^\circ
  5. E
    149∘149^\circ

Answer

The measure of the obtuse angle formed at the intersection of line QBQB and line kk is 121∘121^\circ.
The measure 121∘121^\circ is correct because ray ADAD bisects ∠CAB\angle CAB, giving ∠PAB=(1.5x−5)∘\angle PAB = (1.5x - 5)^\circ. Since AD⊥BFAD \perp BF, △APB\triangle APB is a right triangle where (1.5x−5)+(x+35)=90(1.5x - 5) + (x + 35) = 90, yielding x=24x = 24. Consequently, ∠QAB=62∘\angle QAB = 62^\circ and ∠ABQ=59∘\angle ABQ = 59^\circ. In △ABQ\triangle ABQ, the acute angle ∠AQB=180∘−(62∘+59∘)=59∘\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 59^\circ. The supplementary obtuse angle along line kk is 180∘−59∘=121∘180^\circ - 59^\circ = 121^\circ.

Step-by-Step Solution

1
Express the angle ∠PAB\angle PAB in terms of xx.
Since ray ADAD bisects ∠CAB=(3x−10)∘\angle CAB = (3x - 10)^\circ, ∠PAB=3x−102=(1.5x−5)∘\angle PAB = \frac{3x - 10}{2} = (1.5x - 5)^\circ.
An angle bisector divides an angle into two equal halves.
2
Set up an equation using triangle △APB\triangle APB.
In △APB\triangle APB, ∠APB=90∘\angle APB = 90^\circ because ray AD⊥AD \perp ray BFBF. Therefore, ∠PAB+∠ABP=90∘  ⟹  (1.5x−5)+(x+35)=90  ⟹  2.5x+30=90  ⟹  2.5x=60  ⟹  x=24\angle PAB + \angle ABP = 90^\circ \implies (1.5x - 5) + (x + 35) = 90 \implies 2.5x + 30 = 90 \implies 2.5x = 60 \implies x = 24.
The acute angles in a right triangle sum to 90∘90^\circ.
3
Calculate the measures of ∠CAB\angle CAB and ∠ABQ\angle ABQ.
∠CAB=3(24)−10=62∘\angle CAB = 3(24) - 10 = 62^\circ, and ∠ABQ=∠ABF=24+35=59∘\angle ABQ = \angle ABF = 24 + 35 = 59^\circ.
Substitute x=24x = 24 back into the original angle expressions.
4
Determine the acute angle ∠AQB\angle AQB in triangle △ABQ\triangle ABQ.
In △ABQ\triangle ABQ, points Q,C,AQ, C, A lie on line kk, so ∠QAB=∠CAB=62∘\angle QAB = \angle CAB = 62^\circ. Sum of angles in △ABQ\triangle ABQ: ∠AQB=180∘−(62∘+59∘)=180∘−121∘=59∘\angle AQB = 180^\circ - (62^\circ + 59^\circ) = 180^\circ - 121^\circ = 59^\circ.
The interior angles of any triangle sum to 180∘180^\circ.
5
Find the supplementary obtuse angle at intersection point QQ.
Obtuse angle =180∘−59∘=121∘= 180^\circ - 59^\circ = 121^\circ.
Angles forming a linear pair on a straight line are supplementary.

Key Concept

Parallel Lines, Transversals, Angle Bisectors, and Triangle Angle Sum Theorem
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Topics

Lines and Angles

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