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Angles and Their Measure: 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 Name the opening

Rays $\overrightarrow{JV}$ and $\overrightarrow{JW}$ form a selected angle. Name its vertex and write two valid three-letter names for the angle.
1. Hint 1
  Both rays begin at the same point.
2. Hint 2
  Put the common starting point between the other two letters.
Answer

The vertex is $J$ ; the names are $\angle VJW$ and $\angle WJV$ .

Full solution

Both rays start at $J$ , so that point is the vertex.

The vertex goes in the middle, giving $\angle VJW$ or $\angle WJV$ for the selected opening.

Answer

The vertex is $J$ ; the names are $\angle VJW$ and $\angle WJV$ .

Key idea

The middle letter of an angle name identifies its vertex.
Problem 2 Explain the boundaries

Classify each selected opening:

Case A: $18^\circ$ .

Case B: $90^\circ$ .

Case C: $152^\circ$ .

Case D: $180^\circ$ .

Case E: $271^\circ$ .

Case F: $360^\circ$ .

Use the definitions to explain why $90^\circ$ is neither acute nor obtuse and why $180^\circ$ is neither obtuse nor reflex.
1. Hint 1
  Compare the measures with quarter, half, and full turns.
2. Hint 2
  Apply the strict less-than and greater-than conditions when checking the boundary values.
Answer

In order: acute, right, obtuse, straight, reflex, full turn. The strict bounds exclude $90^\circ$ from acute and obtuse angles, and exclude $180^\circ$ from obtuse and reflex angles.

Full solution

Case A: The measure $18^\circ$ is between zero and $90^\circ$ , so it is acute.

Case B: $90^\circ$ is right.

Case C: The measure $152^\circ$ is between $90^\circ$ and $180^\circ$ , so it is obtuse.

Case D: $180^\circ$ is straight.

Case E: The measure $271^\circ$ is between $180^\circ$ and $360^\circ$ , so it is reflex.

Case F: $360^\circ$ is a full turn.

An acute angle must be less than $90^\circ$ and an obtuse angle must be greater than $90^\circ$ , so neither definition includes $90^\circ$ .

An obtuse angle must be less than $180^\circ$ and a reflex angle must be greater than $180^\circ$ , so neither definition includes $180^\circ$ .

Answer

In order: acute, right, obtuse, straight, reflex, full turn. The strict bounds exclude $90^\circ$ from acute and obtuse angles, and exclude $180^\circ$ from obtuse and reflex angles.

Key idea

Boundary measures form separate classes because the neighboring definitions use strict inequalities.
Problem 3 Select a scale

A protractor is centered on vertex $K$ , and its baseline is aligned with ray $KL$ .

Ray $KL$ meets the outer zero.

Ray $KM$ meets outer 124 and inner 56.

Give the smaller opening's measure and explain your choice of scale.
1. Hint 1
  Choose a scale using the starting ray rather than the ending ray.
2. Hint 2
  Follow the scale from the zero on $KL$ to the mark at $KM$ .
Answer

$m\angle LKM=124^\circ$ ; use the outer scale because its zero lies on $KL$ .

Full solution

The outer zero lies on the starting ray $KL$ .

Following that scale to $KM$ gives
$m\angle LKM=124^\circ.$

The inner reading starts from the opposite direction and measures a different opening.

Answer

$m\angle LKM=124^\circ$ ; use the outer scale because its zero lies on $KL$ .

Key idea

Read the same protractor scale from its starting zero to the second ray.
Problem 4 Draw and check

On paper, draw a ray $PQ$ .

Use a protractor to draw ray $PR$ so that $m\angle QPR=115^\circ$ .

Describe where you place the tool and how you choose the scale.

State the angle type.
1. Hint 1
  The center mark and the zero each have a different alignment job.
2. Hint 2
  After marking the chosen degree, join that mark to the vertex.
Answer

Place the center at $P$ , align the baseline with $PQ$ , mark 115 on the scale starting at $PQ$ , and draw $PR$ through the mark. The angle is obtuse.

Full solution

Draw $PQ$ and place the protractor's center on $P$ .

Align the baseline with $PQ$ , then mark 115 on the scale whose zero lies on $PQ$ .

Remove the protractor and draw $PR$ through that mark.

The measure is between $90^\circ$ and $180^\circ$ , so the opening is obtuse; the paper measurement is approximate.

Answer

Place the center at $P$ , align the baseline with $PQ$ , mark 115 on the scale starting at $PQ$ , and draw $PR$ through the mark. The angle is obtuse.

Key idea

A measured drawing needs both center alignment and the correct starting zero.
Problem 5 Complete an opening

Ray $OB$ lies inside the selected angle $AOC$ . The whole measures $161^\circ$ , and $m\angle AOB=73^\circ$ . Find $m\angle BOC$ .
1. Hint 1
  The interior ray divides the whole opening into two parts.
2. Hint 2
  Subtract the known part from the whole.
Answer

$m\angle BOC=88^\circ$ .

Full solution

Angle addition gives
$73^\circ+m\angle BOC=161^\circ.$

Subtracting the known part gives
$161^\circ-73^\circ=88^\circ.$

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

$m\angle BOC=88^\circ$ .

Key idea

An unknown angular part is the whole opening minus the known part.
Problem 6 The other way around

Two rays bound a selected reflex opening of $252^\circ$ . Find their smaller opening and the measure of each part when that smaller opening is bisected.
1. Hint 1
  The two openings together complete a full turn.
2. Hint 2
  Find the smaller opening before halving it.
Answer

The smaller opening is $108^\circ$ ; each bisected part is $54^\circ$ .

Full solution

The smaller opening measures
$360^\circ-252^\circ=108^\circ.$

The bisector divides that opening equally, so each part measures
$108^\circ\div2=54^\circ.$

Answer

The smaller opening is $108^\circ$ ; each bisected part is $54^\circ$ .

Key idea

Select the intended opening before applying an angle bisector.
Problem 7 Three consecutive parts

Three nonoverlapping consecutive angles fill one selected $198^\circ$ opening.

Their measures, in order, are $(x+20)^\circ$ , $(2x+10)^\circ$ , and $(x+8)^\circ$ .

Find $x$ and all three measures.
1. Hint 1
  The three part measures must add to the specified whole.
2. Hint 2
  Combine the variable terms and constants, then substitute your value into each part.
Answer

$x=40$ ; the measures are $60^\circ$ , $90^\circ$ , and $48^\circ$ , in order.

Full solution

Angle addition gives $4x+38=198$ .

Subtracting 38 and dividing by 4 gives $x=40$ .

Substitution gives $60^\circ$ , $90^\circ$ , and $48^\circ$ .

Their sum is $198^\circ$ , as required by the selected reflex opening.

Answer

$x=40$ ; the measures are $60^\circ$ , $90^\circ$ , and $48^\circ$ , in order.

Key idea

Angle addition applies around a specified reflex opening as well as a smaller opening.
Problem 8 Keep track of overlap

Ray $OB$ lies inside the selected angle $AOC$ , with $m\angle AOB=47^\circ$ and $m\angle BOC=62^\circ$ .

Find $m\angle AOC$ .

Then calculate $m\angle AOC+m\angle BOC$ and explain which region that sum counts twice.
1. Hint 1
  First add the two parts that fill $AOC$ .
2. Hint 2
  The whole opening already contains the part $BOC$ .
Answer

$m\angle AOC=109^\circ$ ; the second sum is $171^\circ$ and counts the $62^\circ$ region $BOC$ twice.

Full solution

The nonoverlapping parts add to
$47^\circ+62^\circ=109^\circ.$

Adding $BOC$ again gives
$109^\circ+62^\circ=171^\circ.$

The whole $AOC$ already contains $BOC$ , so this second sum counts that part twice rather than measuring the whole once.

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

$m\angle AOC=109^\circ$ ; the second sum is $171^\circ$ and counts the $62^\circ$ region $BOC$ twice.

Key idea

A shared vertex does not guarantee that a sum of angle measures counts a region once.
Problem 9 Equal parts with expressions

Ray $HT$ bisects the selected angle $RHS$ . Its parts measure $(4x+3)^\circ$ and $(6x-21)^\circ$ . Find $x$ , each part, and the whole measure.
1. Hint 1
  A bisector gives an equality between the two part measures.
2. Hint 2
  Solve that equality before adding the two equal parts.
Answer

$x=12$ ; each part is $51^\circ$ and the whole is $102^\circ$ .

Full solution

Equal parts give
$4x+3=6x-21.$

Rearranging gives $24=2x$ , so $x=12$ .

Each part measures $4(12)+3=51$ degrees.

Angle addition gives
$51^\circ+51^\circ=102^\circ.$

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

$x=12$ ; each part is $51^\circ$ and the whole is $102^\circ$ .

Key idea

The bisector supplies equality; angle addition supplies the whole measure.
Problem 10 Choose the reflex route

The selected reflex opening from ray $OA$ to ray $OB$ measures $244^\circ$ .

Ray $OC$ bisects that opening.

Ray $OD$ lies inside its selected part $AOC$ , with $m\angle AOD=49^\circ$ .

Find $m\angle DOC$ and $m\angle COB$ , and classify both of those smaller angles.
1. Hint 1
  Find each half of the selected reflex opening first.
2. Hint 2
  Subtract $AOD$ from the half $AOC$ ; the other half is unchanged.
Answer

$m\angle DOC=73^\circ$ , acute; $m\angle COB=122^\circ$ , obtuse.

Full solution

The bisector makes each half measure
$244^\circ\div2=122^\circ.$

Within the selected half $AOC$ , angle addition gives
$122^\circ-49^\circ=73^\circ.$

Thus $DOC$ is acute, while the other half $COB$ remains $122^\circ$ and is obtuse.

The blue first ray and dashed red final ray bound the selected counterclockwise opening. Any named interior rays separate the consecutive parts. Use the stated angle measures and markings, rather than measuring the drawing.

Answer

$m\angle DOC=73^\circ$ , acute; $m\angle COB=122^\circ$ , obtuse.

Key idea

Track which selected opening is being halved and which part is being subtracted.

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