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Angle Pairs: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The shared side

Ray $OB$ lies inside the non-reflex angle $AOC$ . The consecutive angles $AOB$ and $BOC$ measure $28^\circ$ and $62^\circ$ . Describe the pair as adjacent or nonadjacent, and as complementary, supplementary, or neither. Find and classify $\angle AOC$ .

The labels give the two nonoverlapping parts of the selected opening.

Text description of this figure

Rays OA, OB, and OC have common vertex O. Ray OB is inside angle AOC. The two consecutive sectors are labeled 28 degrees and 62 degrees.
1. Hint 1
  Use the shared ray and the stated interior position to test adjacency.
2. Hint 2
  Compare the sum of the two given measures with the special angle totals.
Answer

The pair is adjacent and complementary; $m\angle AOC=90^\circ$ , so $\angle AOC$ is right.

Full solution

The angles share vertex O and side OB.
Since OB is inside the selected opening, their interiors do not overlap, so they are adjacent.

Angle addition gives $m\angle AOC=28^\circ+62^\circ$ , hence
$m\angle AOC=90^\circ.$
The pair is complementary because its measure sum is $90^\circ$ , and the whole angle is right.

Answer

The pair is adjacent and complementary; $m\angle AOC=90^\circ$ , so $\angle AOC$ is right.

Key idea

Adjacency concerns the rays' positions; complementarity concerns the sum of the measures.
Problem 2 The opposite outer rays

Points $L,P,N$ are collinear in that order. Ray $PM$ is not on their line, and $m\angle LPM=157^\circ$ . Find $m\angle MPN$ and justify the equation you use.
1. Hint 1
  Translate the point order into a statement about the two outer rays.
2. Hint 2
  The shared ray divides the selected straight opening into two parts.
Answer

$m\angle MPN=23^\circ$ .

Full solution

Because P is between L and N, rays PL and PN are opposite.
The two named angles share PM and have disjoint interiors, so they form a linear pair.

Their measures therefore total $180^\circ$ .
Subtracting the known measure gives $m\angle MPN=180^\circ-157^\circ$ , so
$m\angle MPN=23^\circ.$
The measures sum to $180^\circ$ as required.

Answer

$m\angle MPN=23^\circ$ .

Key idea

Betweenness and a shared ray justify the straight-angle sum.
Problem 3 All four sectors

Two distinct lines cross at $O$ . The northwest (NW) sector measures $97^\circ$ . Find the other three sector measures, listing them clockwise from NW as northeast (NE), southeast (SE), and southwest (SW). Give a reason for each measure.

The sector labels record the supplied information.

Text description of this figure

Two distinct lines cross at O. The northwest sector is labeled 97 degrees. The northeast, southeast, and southwest sectors each have a question mark. No angle equality marks are given.
1. Hint 1
  Separate neighboring sectors from opposite sectors.
2. Hint 2
  Use a straight-angle total for a neighbor and congruence for an opposite sector.
Answer

NE $83^\circ$ ; SE $97^\circ$ ; SW $83^\circ$ .

Full solution

The NE sector forms a linear pair with the NW sector, so it measures
$180^\circ-97^\circ=83^\circ.$

The SE sector is vertical to the NW sector, so its measure is $97^\circ$ .
The SW sector is vertical to the NE sector, so its measure is $83^\circ$ .

Checking the turn gives $97+83+97+83=360$ degrees.
Each neighboring pair also totals $180^\circ$ .

Answer

NE $83^\circ$ ; SE $97^\circ$ ; SW $83^\circ$ .

Key idea

At two distinct intersecting lines, opposite sectors agree and neighboring sectors are supplementary.
Problem 4 Equal across the vertex

Two distinct lines intersect. A pair of vertical angles measures $(9x-22)^\circ$ and $(6x+23)^\circ$ . Find $x$ , the measure of each angle in that pair, and the measure of each remaining sector.

The sector labels record the supplied information.

Text description of this figure

Two distinct lines cross at O. The top, right, bottom, and left sectors are labeled 9x - 22, with no value, 6x + 23, and with no value, respectively. No angle equality marks are given.
1. Hint 1
  Determine whether the given measures should be equal or should be added to a fixed total.
2. Hint 2
  After solving for x, substitute into both expressions before finding a neighboring sector.
Answer

$x=15$ ; each named vertical angle measures $113^\circ$ ; each remaining sector measures $67^\circ$ .

Full solution

Vertical angles are congruent, so
$9x-22=6x+23.$
Subtracting $6x$ and adding 22 gives $3x=45$ .
Dividing by 3 gives $x=15$ .

Substitution gives $9(15)-22=113$ and
$6(15)+23=113.$
Thus each named angle measures $113^\circ$ .

Each remaining sector forms a linear pair with one of those angles.
Its measure is
$180^\circ-113^\circ=67^\circ.$

Answer

$x=15$ ; each named vertical angle measures $113^\circ$ ; each remaining sector measures $67^\circ$ .

Key idea

The geometric relationship determines the equation before algebra determines the measures.
Problem 5 Equal partners in different places

Angles $U$ and $V$ are congruent acute angles. Angle $A$ complements $U$ , and angle $B$ complements $V$ . If $m\angle A=(3x+8)^\circ$ and $m\angle B=(5x-34)^\circ$ , find $x$ and the measures of $A$ , $B$ , $U$ , and $V$ .
1. Hint 1
  Compare the totals from which the congruent acute angles are removed.
2. Hint 2
  Write an equality for the two complement measures, then subtract each from the complement total.
Answer

$x=21$ ; $m\angle A=m\angle B=71^\circ$ ; $m\angle U=m\angle V=19^\circ$ .

Full solution

Complements of congruent angles are congruent, so
$3x+8=5x-34.$
Subtracting $3x$ and adding 34 gives $42=2x$ , hence $x=21$ .

The two complement expressions give $3(21)+8=71$ and
$5(21)-34=71.$
Thus both A and B measure $71^\circ$ .

Each acute angle being complemented measures
$90^\circ-71^\circ=19^\circ.$
This verifies both the congruence and the required acute measures of U and V.

Answer

$x=21$ ; $m\angle A=m\angle B=71^\circ$ ; $m\angle U=m\angle V=19^\circ$ .

Key idea

Equal angles removed from the same total leave equal partner measures.
Problem 6 One circuit of the vertex

Four distinct rays occur in the cyclic order $OA,OB,OC,OD$ around $O$ . Their consecutive sectors fill a complete turn exactly once and measure $48^\circ$ , $87^\circ$ , $(3x)^\circ$ , and $(2x+10)^\circ$ , respectively. Find $x$ and the two unknown sector measures.

Every sector in the complete turn is represented once.

Text description of this figure

Four distinct rays from O occur in the stated cyclic order. Successive sectors carry labels 48 degrees, 87 degrees, 3x degrees, and 2x plus 10 degrees. They fill one full turn.
1. Hint 1
  Identify the whole that the four sectors fill.
2. Hint 2
  Add all four expressions, combine like terms, and then check the resulting sectors.
Answer

$x=43$ ; $m\angle COD=129^\circ$ and $m\angle DOA=96^\circ$ .

Full solution

The sectors fill a full turn, so
$48+87+3x+(2x+10)=360.$
Combining terms gives
$5x+145=360.$

Subtracting 145 gives $5x=215$ .
Dividing by 5 gives $x=43$ .

The unknown measures are $3(43)^\circ=129^\circ$ and
$(2(43)+10)^\circ=96^\circ.$
All sectors are positive and non-reflex, and $48+87+129+96=360$ verifies the whole turn.

Answer

$x=43$ ; $m\angle COD=129^\circ$ and $m\angle DOA=96^\circ$ .

Key idea

A full-turn equation needs every consecutive sector, with no missing or repeated part.
Problem 7 A bisected part of a straight opening

Points $A,O,C$ are collinear in that order. Ray $OB$ is not on the line, and $OD$ bisects $\angle BOC$ . Given $m\angle AOB=(4m+12)^\circ$ and $m\angle BOD=(m+15)^\circ$ , find $m$ and $m\angle BOC$ .

The entire straight opening includes AOB and both halves of BOC.

Text description of this figure

A, O, C are collinear in that order. Ray OB is on one side of the line, and OD is inside angle BOC. Labels show the given expressions on AOB and BOD. The bisector condition is stated in the problem.
1. Hint 1
  Which expression describes half of BOC? Use it to express the whole BOC.
2. Hint 2
  Add AOB and the whole BOC to fill the straight opening.
Answer

$m=23$ and $m\angle BOC=76^\circ$ .

Full solution

Since OD bisects angle BOC, both halves have measure $(m+15)^\circ$ .
Angle addition gives
$m\angle BOC=2(m+15)^\circ.$

Angles AOB and BOC form a linear pair, so
$4m+12+2(m+15)=180.$
Expanding gives $6m+42=180$ , so $6m=138$ and $m=23$ .

Each half of BOC measures
$(23+15)^\circ=38^\circ.$
Thus
$m\angle BOC=76^\circ.$
The other part measures $104^\circ$ , and
$104^\circ+76^\circ=180^\circ.$

Answer

$m=23$ and $m\angle BOC=76^\circ$ .

Key idea

Keep a bisected part distinct from the whole angle when forming an equation.
Problem 8 Test the position claim

Consider the claim that every pair of supplementary angles is a linear pair. Draw a counterexample using angles with different vertices and measures $98^\circ$ and $82^\circ$ . Explain why your drawing meets the hypothesis but fails the conclusion.
1. Hint 1
  Check the measure requirement separately from the position requirement.
2. Hint 2
  Use the different vertices to test a necessary condition for a linear pair.
Answer

The claim is false; separate angles of 98 degrees and 82 degrees are supplementary and do not form a linear pair.

Full solution

Draw a $98^\circ$ angle at P and an $82^\circ$ angle at a different point Q.
Their measure sum is $98^\circ+82^\circ=180^\circ$ , so they satisfy the supplementary hypothesis.

The angles have different vertices, so they are not adjacent.
A linear pair must be adjacent with opposite noncommon rays.
The conclusion therefore fails for this pair, which refutes the universal claim.

The measure sum holds while the linear-pair position requirement fails.

Answer

The claim is false; separate angles of 98 degrees and 82 degrees are supplementary and do not form a linear pair.

Key idea

A single counterexample can separate a measure condition from a stronger geometric configuration.
Problem 9 Two overlapping larger openings

Points $A,O,D$ are collinear in that order. Rays $OB$ and $OC$ split one straight opening into the consecutive sectors $AOB$ , $BOC$ , and $COD$ . Suppose $\angle AOB\cong\angle COD$ and $m\angle BOC=48^\circ$ . Find the measures of $AOB$ , $COD$ , $AOC$ , and $BOD$ , and justify why the last two angles are congruent.

Each larger requested opening contains the middle sector and one of the congruent outer sectors.

Text description of this figure

The straight opening AOD is divided into sectors AOB, BOC, and COD in that order. BOC has measure 48 degrees. Single arcs on AOB and COD record the given congruence.
1. Hint 1
  The two outer sectors have equal measures and share the remaining part of a straight-angle total.
2. Hint 2
  Build each larger opening from one outer sector and the same middle sector.
Answer

$m\angle AOB=m\angle COD=66^\circ$ ; $m\angle AOC=m\angle BOD=114^\circ$ , so $\angle AOC\cong\angle BOD$ .

Full solution

Removing the $48^\circ$ middle sector from the straight opening leaves $180^\circ-48^\circ=132^\circ$ for the two congruent outer sectors.
Each therefore measures
$132^\circ\div2=66^\circ.$

Angle addition gives
$m\angle AOC=66^\circ+48^\circ=114^\circ.$
It also gives
$m\angle BOD=48^\circ+66^\circ=114^\circ.$

The two larger angles have equal measures, so they are congruent.
This conclusion follows from adding the same middle measure to equal outer measures, although the larger angles have overlapping interiors.

Answer

$m\angle AOB=m\angle COD=66^\circ$ ; $m\angle AOC=m\angle BOD=114^\circ$ , so $\angle AOC\cong\angle BOD$ .

Key idea

Equal angle parts can produce equal larger openings even when those larger openings overlap.
Problem 10 Design a crossing under two conditions

Case A: Draw two distinct intersecting lines whose one pair of vertical angles is also complementary. Label all four sector measures and justify that your configuration satisfies both requirements.

Case B: Could two adjacent sectors at the crossing of two distinct lines be complementary? Either construct such a crossing or justify why it is impossible.
1. Hint 1
  Case A: impose both the position rule for vertical angles and the measure rule for complementary angles.
2. Hint 2
  Case A: once that pair is fixed, use a neighboring sector to complete the crossing.
3. Hint 3
  Case B: determine the sum already forced by adjacency at a crossing of two full lines.
Answer

Case A: opposite sectors $45^\circ$ and $45^\circ$ , with the other opposite pair $135^\circ$ and $135^\circ$ . Case B: impossible.

Full solution

Case A: vertical angles are congruent, and the requested pair must total $90^\circ$ .
Each therefore measures $45^\circ$ .
Draw a line through O, draw a second line making a $45^\circ$ opening with it, and extend both lines in both directions.

Case A: the vertical opposite of the $45^\circ$ sector is also $45^\circ$ .
Each neighboring sector measures
$180^\circ-45^\circ=135^\circ.$
The selected opposite pair is vertical and complementary, and all four sectors sum to $360^\circ$ .

Case B: two adjacent sectors at two distinct intersecting lines have opposite noncommon rays.
They form a linear pair, so their measure sum is $180^\circ$ .
Complementarity would require the same sum to be $90^\circ$ , which is impossible.

Case A can be realized by crossing the lines at a 45-degree angle.

Answer

Case A: opposite sectors $45^\circ$ and $45^\circ$ , with the other opposite pair $135^\circ$ and $135^\circ$ . Case B: impossible.

Key idea

A proposed configuration must satisfy its position and measure requirements at the same time.

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