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The Exterior Angle Theorem: 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 angles from the extension

Triangle $GHK$ has $HK$ extended beyond $K$ to $L$ , with $H,K,L$ in that order. Name the exterior angle at $K$ , its adjacent interior angle, and its two remote interior angles.

Use the stated point order to identify the angle roles.

Text description of this figure

Triangle GHK has vertices G, H and K. Side HK continues on the same line beyond K to L. Arcs show angle positions without names or measures.
1. Hint 1
  Locate the vertex where the side continues beyond the triangle.
2. Hint 2
  The remote interior angles are the ones at the other two triangle vertices.
Answer

Exterior $\angle GKL$ ; adjacent interior $\angle GKH$ ; remote interiors $\angle HGK$ and $\angle GHK$ (reversing the first and third letters of any angle name is equivalent).

Full solution

The continued side ends at $K$ and runs toward $L$ , so $\angle GKL$ is the exterior angle.
It uses the triangle side $KG$ and the extended ray $KL$ .

The interior angle at the same vertex is $\angle GKH$ .
Rays $KH$ and $KL$ are opposite, so this interior angle and the exterior angle form a linear pair.

The remote interior angles are at the other vertices, $G$ and $H$ .
Their names are $\angle HGK$ and $\angle GHK$ .
Each name keeps its vertex in the middle.

Answer

Exterior $\angle GKL$ ; adjacent interior $\angle GKH$ ; remote interiors $\angle HGK$ and $\angle GHK$ (reversing the first and third letters of any angle name is equivalent).

Key idea

A chosen exterior angle fixes one adjacent interior angle and two remote interior angles.
Problem 2 Two remote measures

Triangle $ABC$ has $BC$ extended beyond $C$ to $D$ , with $B,C,D$ in that order. Given $m\angle BAC=29^\circ$ and $m\angle ABC=76^\circ$ , find $m\angle ACD$ and $m\angle ACB$ . State the relationship used for each.
1. Hint 1
  First separate the remote interior angles from the adjacent interior angle.
2. Hint 2
  The exterior measure equals the sum of the remote measures; its adjacent interior measure completes a straight angle.
Answer

$m\angle ACD=105^\circ$ , by the exterior angle theorem; $m\angle ACB=75^\circ$ , by the linear-pair rule.

Full solution

The interior angles at $A$ and $B$ are remote for $\angle ACD$ .
The exterior angle theorem gives
$m\angle ACD=29^\circ+76^\circ$ $m\angle ACD=105^\circ$

The adjacent interior angle $\angle ACB$ forms a linear pair with the exterior angle because $CB$ and $CD$ are opposite rays.
Therefore
$m\angle ACB=180^\circ-105^\circ$ $m\angle ACB=75^\circ$

The interior check is
$29^\circ+76^\circ+75^\circ=180^\circ.$

The remote sum and the adjacent supplement use different angle pairs.

Answer

$m\angle ACD=105^\circ$ , by the exterior angle theorem; $m\angle ACB=75^\circ$ , by the linear-pair rule.

Key idea

Add the remote angles to find an exterior angle, then subtract from a straight angle for its neighbor.
Problem 3 Work backward from the exterior

Triangle $UVW$ has $VW$ extended past $W$ to $X$ , with $V,W,X$ in that order. Given $m\angle UWX=139^\circ$ and $m\angle VUW=82^\circ$ , find $m\angle UVW$ and $m\angle UWV$ .

The labels record the givens; the sketch is not drawn to scale.

Text description of this figure

Triangle UVW has VW extended beyond W to X, with V, W, X on one line in that order. Angle UWX is labeled 139 degrees and angle VUW is labeled 82 degrees. No other measures are shown.
1. Hint 1
  The two requested angles have different roles relative to the given exterior angle.
2. Hint 2
  Subtract the known remote measure from the exterior measure for the other remote; use a linear pair for the interior at W.
Answer

$m\angle UVW=57^\circ$ and $m\angle UWV=41^\circ$ .

Full solution

The exterior angle at $W$ equals the sum of the interior measures at $U$ and $V$ .
Thus
$139^\circ=82^\circ+m\angle UVW$
Subtracting $82^\circ$ gives
$m\angle UVW=57^\circ.$

The interior angle at $W$ is adjacent to the exterior angle and forms a linear pair with it.
Thus
$m\angle UWV=180^\circ-139^\circ$ $m\angle UWV=41^\circ$

The three interior measures $82^\circ$ , $57^\circ$ , and $41^\circ$ total $180^\circ$ .

Answer

$m\angle UVW=57^\circ$ and $m\angle UWV=41^\circ$ .

Key idea

A remote angle is the difference between an exterior angle and the other remote angle.
Problem 4 Expressions attached to angle roles

Triangle $ABC$ has $BC$ extended beyond $C$ to $D$ , with $B,C,D$ in that order. The degree measures are $m\angle BAC=6x-1$ , $m\angle ABC=2x+12$ , and $m\angle ACD=11x-4$ . Find $x$ and all three interior measures. Check the exterior relationship.

The labels record the givens; the sketch is not drawn to scale.

Text description of this figure

Triangle ABC has BC continued beyond C to D. The interior angles at A and B are labeled 6x minus 1 and 2x plus 12 degrees. The exterior angle at C is labeled 11x minus 4 degrees.
1. Hint 1
  Use the relationship determined by the angle positions, rather than adding all three displayed expressions.
2. Hint 2
  Set the sum of the two remote expressions equal to the exterior expression.
3. Hint 3
  After solving for x, the unlisted interior measure is the supplement of the exterior measure.
Answer

$x=5$ ; interior measures $m\angle BAC=29^\circ$ , $m\angle ABC=22^\circ$ , and $m\angle ACB=129^\circ$ ; exterior measure $51^\circ$ .

Full solution

The angles at $A$ and $B$ are remote for the exterior angle at $C$ .
Their measures give
$(6x-1)+(2x+12)=11x-4$

Combining terms gives
$8x+11=11x-4.$
Subtracting $8x$ and adding 4 gives $15=3x$ , hence $x=5$ .

Substituting gives $m\angle BAC=29^\circ$ , $m\angle ABC=22^\circ$ , and
$m\angle ACD=51^\circ.$
The adjacent interior measure is
$m\angle ACB=180^\circ-51^\circ$ $m\angle ACB=129^\circ$

The exterior check is
$29^\circ+22^\circ=51^\circ.$
The interior check is $29^\circ+22^\circ+129^\circ=180^\circ$ , with every interior measure positive.

Answer

$x=5$ ; interior measures $m\angle BAC=29^\circ$ , $m\angle ABC=22^\circ$ , and $m\angle ACB=129^\circ$ ; exterior measure $51^\circ$ .

Key idea

Translate the exterior angle theorem first, then check the algebraic solution against the geometric conditions.
Problem 5 Exterior information at two vertices

Triangle $ABC$ has a non-reflex exterior angle of $132^\circ$ at $A$ and one of $143^\circ$ at $B$ . Find all three interior angle measures and a non-reflex exterior measure at $C$ . Justify why the two exterior choices at either vertex have the same measure.
1. Hint 1
  Each supplied exterior angle is adjacent to the interior angle at its own vertex.
2. Hint 2
  Find the interior measures at A and B before applying the triangle sum.
3. Hint 3
  Either incident side extension makes a linear pair with the same interior angle.
Answer

Interior measures at $A,B,C$ are $48^\circ,37^\circ,95^\circ$ ; an exterior at $C$ measures $85^\circ$ . The two exterior choices at any one vertex have equal measures.

Full solution

The exterior and interior angles at each given vertex form a linear pair.
Thus the interior measure at $A$ is $180^\circ-132^\circ=48^\circ$ , and the one at $B$ is
$180^\circ-143^\circ=37^\circ.$

The triangle sum gives
$m\angle ACB=180^\circ-48^\circ-37^\circ$ $m\angle ACB=95^\circ$

An exterior angle at $C$ measures
$180^\circ-95^\circ=85^\circ.$
This also equals the remote sum $48^\circ+37^\circ$ .

At any one vertex, either incident side can be extended beyond that vertex.
Each resulting non-reflex exterior angle supplements the same interior angle, so their measures are equal.

Answer

Interior measures at $A,B,C$ are $48^\circ,37^\circ,95^\circ$ ; an exterior at $C$ measures $85^\circ$ . The two exterior choices at any one vertex have equal measures.

Key idea

Either side extension gives the same non-reflex exterior measure at a chosen vertex.
Problem 6 An exterior opening split into parts

Triangle $ABC$ has $BC$ extended beyond $C$ to $D$ , with $B,C,D$ in that order. The remote interior measures are $m\angle BAC=46^\circ$ and $m\angle ABC=81^\circ$ . Ray $CE$ lies strictly inside $\angle ACD$ , and $m\angle ECD=33^\circ$ . Find $m\angle ACE$ and $m\angle ACB$ .

Ray CE divides the exterior opening into two adjacent parts.

Text description of this figure

Triangle ABC has BC extended beyond C to D. The interior angles at A and B are given as 46 and 81 degrees. Ray CE lies inside exterior angle ACD, splitting off angle ECD of 33 degrees. No derived measures are shown.
1. Hint 1
  Distinguish the whole exterior angle from the part split off by CE.
2. Hint 2
  Find the whole exterior measure from the two remote interior measures.
3. Hint 3
  Use angle addition for the two exterior parts, and a linear pair for the adjacent interior angle.
Answer

$m\angle ACE=94^\circ$ and $m\angle ACB=53^\circ$ .

Full solution

The exterior angle theorem uses the two remote interior angles.
It gives
$m\angle ACD=46^\circ+81^\circ$ $m\angle ACD=127^\circ$

Because ray $CE$ lies inside $\angle ACD$ , the parts $\angle ACE$ and $\angle ECD$ fill it without overlap.
Angle addition gives
$m\angle ACE=127^\circ-33^\circ$ $m\angle ACE=94^\circ$

The interior angle at $C$ supplements the whole exterior angle, not just its $33^\circ$ part.
Thus
$m\angle ACB=180^\circ-127^\circ$ $m\angle ACB=53^\circ$

Answer

$m\angle ACE=94^\circ$ and $m\angle ACB=53^\circ$ .

Key idea

Find the whole exterior angle before using an interior ray to subtract one of its parts.
Problem 7 A triangular robot route

A robot follows three straight sides of a triangle once and turns right through a non-reflex exterior angle at each corner. Its first two turns measure $108^\circ$ and $146^\circ$ . Find its third turn and the triangle's interior angle at that third corner. Explain why adding the turns is appropriate.
1. Hint 1
  Track what changes when the robot reaches each corner: its direction.
2. Hint 2
  All three exterior turns follow the same boundary traversal and the same turning direction.
3. Hint 3
  The exterior turns total a full turn; the interior at a corner supplements that corner's exterior turn.
Answer

Third turn $106^\circ$ ; third interior angle $74^\circ$ ; the three right turns total $360^\circ$ .

Full solution

The robot follows the triangle once and takes one exterior turn at every vertex, all to the right.
Thus the turn measures total $360^\circ$ .
The third turn is
$360^\circ-108^\circ-146^\circ=106^\circ$

The adjacent interior angle completes a linear pair with that exterior turn.
Its measure is
$180^\circ-106^\circ=74^\circ$

The other interior measures are $72^\circ$ and $34^\circ$ .
Their sum with $74^\circ$ is $180^\circ$ , checking the triangle data.
A consistent direction is what makes the three exterior measures a record of the robot's turns.

Answer

Third turn $106^\circ$ ; third interior angle $74^\circ$ ; the three right turns total $360^\circ$ .

Key idea

A full triangular traversal uses three exterior turns in one direction, totaling one full turn.
Problem 8 Locate and repair the first invalid line

Triangle $ABC$ has $BC$ extended beyond $C$ to $D$ , with $B,C,D$ in that order. Let $a=m\angle BAC$ , $b=m\angle ABC$ , $c=m\angle ACB$ , and $e=m\angle ACD$ . A proposed argument contains these lines.

Line 1. $a+b+c=180^\circ$ .

Line 2. $c+e=180^\circ$ .

Line 3. $e=180^\circ-(a+b)$ .

Identify the first invalid line, replace it with a correct equation that follows from the preceding lines, and use $a=47^\circ$ , $b=72^\circ$ to find $c$ and $e$ .
1. Hint 1
  Check each line against its stated angle roles before looking at the numerical data.
2. Hint 2
  Both left sides in Lines 1 and 2 equal the same straight-angle measure.
3. Hint 3
  Set the left sides equal and subtract the common measure c.
Answer

Line 3 is the first invalid line; replace it with $e=a+b$ . The measures are $c=61^\circ$ and $e=119^\circ$ .

Full solution

Line 1 is valid by the triangle angle sum.
Line 2 is valid because the stated side extension makes the interior angle $c$ and exterior angle $e$ a linear pair.

Line 3 is the first invalid line.
The expression $180^\circ-(a+b)$ is the remaining interior measure $c$ , not the exterior measure $e$ .

The valid repair equates the preceding totals: $a+b+c=c+e$ .
Subtracting $c$ from both sides gives $e=a+b$ .

Using the data in Line 1 gives
$c=180^\circ-47^\circ-72^\circ$ $c=61^\circ$
The repaired Line 3 gives
$e=47^\circ+72^\circ$ $e=119^\circ$

Finally, $61^\circ+119^\circ=180^\circ$ checks Line 2.

The two correct totals remove the same interior measure c.

Answer

Line 3 is the first invalid line; replace it with $e=a+b$ . The measures are $c=61^\circ$ and $e=119^\circ$ .

Key idea

When two sums equal the same total, cancel their common part rather than replacing one angle by another.
Problem 9 Find the range compatible with a triangle

Triangle $ABC$ has $BC$ extended beyond $C$ to $D$ , with $B,C,D$ in that order. The exterior measure is $(5t+10)^\circ$ , and the remote interior angle at $A$ measures $(2t+49)^\circ$ . Find all real values of $t$ compatible with positive measures for all three interior angles and a non-reflex exterior angle.

For every $t$ in that interval, justify that the exterior angle is greater than each remote interior angle. Explain which exterior-angle comparison is expressed by the lower bound you found.

The labels record the givens; the sketch is not drawn to scale.

Text description of this figure

Triangle ABC has BC continued beyond C to D. The interior angle at A is labeled 2t plus 49 degrees and exterior angle ACD is labeled 5t plus 10 degrees. The other measures are unlabeled.
1. Hint 1
  The unlisted interior angles must also have positive measures.
2. Hint 2
  Express the other remote measure as exterior minus the given remote measure.
3. Hint 3
  Express the adjacent interior measure as 180 degrees minus the exterior measure, then combine the positivity conditions.
4. Hint 4
  An exterior angle equals the sum of its two positive remote interior measures. Compare it with each remote measure by subtracting that measure.
Answer

$13<t<34$ ; for every valid $t$ , the exterior angle is greater than each remote interior angle. The lower bound $t>13$ expresses $m\angle ACD>m\angle BAC$ .

Full solution

By the exterior angle theorem, the other remote interior measure in degrees is
$(5t+10)-(2t+49)=3t-39$
Its positivity gives $3t-39>0$ , so $t>13$ .

The interior angle adjacent to the exterior angle has measure $(170-5t)^\circ$ .
Its positivity gives $170-5t>0$ , hence $t<34$ .

For $13<t<34$ , the supplied remote measure $2t+49$ is positive as well.
The exterior measure $5t+10$ is positive and less than $180$ , as required.

The variable terms in the three interior measures cancel because
$2t+3t-5t=0.$
The constants contribute $49^\circ-39^\circ=10^\circ$ , followed by
$10^\circ+170^\circ=180^\circ.$
Thus all the stated measure conditions hold throughout, and only throughout, the open interval $13<t<34$ .

For every $t$ in the interval, the remote measures $(2t+49)^\circ$ and $(3t-39)^\circ$ are both positive.
The exterior angle theorem makes their sum the exterior measure, so it exceeds each remote measure.

Subtracting the remote measure at $A$ from the exterior measure gives
$(3t-39)^\circ>0^\circ.$
Thus $5t+10>2t+49$ , which simplifies to $t>13$ .
This lower bound is exactly the exterior angle inequality for that remote angle.

Subtracting the other remote measure instead gives
$(2t+49)^\circ>0^\circ.$
It is positive throughout $13<t<34$ , so $5t+10>3t-39$ also holds.
These two strict comparisons establish the exterior angle inequality for each remote angle.

Answer

$13<t<34$ ; for every valid $t$ , the exterior angle is greater than each remote interior angle. The lower bound $t>13$ expresses $m\angle ACD>m\angle BAC$ .

Key idea

Positive remote interior measures both restrict the allowed values and make the exterior angle strictly greater than either remote angle.
Problem 10 Check a directed turning record

Triangle $ABC$ has interior measures $53^\circ$ at $A$ , $61^\circ$ at $B$ , and $66^\circ$ at $C$ . Follow the boundary in the order $B$ to $C$ to $A$ to $B$ , with a left turn at each vertex.

Case A. Record the three non-reflex exterior turn measures at $A,B,C$ and justify their total using interior angles.

Case B. Keep the exterior choices at $A$ and $C$ , but at $B$ replace its left turn by the other exterior angle, formed by extending $BC$ backward beyond $B$ . A record claims that because the three positive measures still total $360^\circ$ , the replacement also describes the same boundary traversal. Decide whether that claim is correct, distinguishing the angle measures from their turn directions.

The arrows specify the boundary order for Case A; the sketch is not drawn to scale.

Text description of this figure

The directed triangular path runs B to C to A to B. At each corner, a dashed continuation of the incoming side and a blue arc identify the left exterior turn, labeled eA, eB or eC. No angle measures or alternative at B are shown.
1. Hint 1
  Case A. Each recorded exterior turn is adjacent to its interior angle.
2. Hint 2
  Case A. Add the three supplements by subtracting the interior-angle total from three straight angles.
3. Hint 3
  Case B. The other exterior angle at B has the same positive measure, but check which traversal makes it a turn from the continued incoming side toward the outgoing side.
Answer

Case A: at $A,B,C$ , the turns are $127^\circ,119^\circ,114^\circ$ , totaling $360^\circ$ . Case B: the claim is false; the sum remains $360^\circ$ , but the replacement reverses the turn direction at $B$ .

Full solution

Case A.
Each exterior turn supplements its adjacent interior angle.
At $A$ the turn is
$180^\circ-53^\circ=127^\circ.$
At $B$ it is
$180^\circ-61^\circ=119^\circ.$
At $C$ it is
$180^\circ-66^\circ=114^\circ.$

Case A.
Adding those three supplements starts with $540^\circ$ and subtracts the triangle's $180^\circ$ interior-angle total.
The result is $360^\circ$ .
Numerically,
$127^\circ+119^\circ+114^\circ=360^\circ.$
The arrows and incoming-side continuations make each of these a left turn along the stated order.

Case B.
Extending the other incident side at $B$ gives another exterior angle that also supplements the same $61^\circ$ interior angle.
Its positive measure is therefore still $119^\circ$ , so the numerical sum does not change.

Case B.
On the stated path, the incoming side at $B$ is $AB$ , and the outgoing side is $BC$ .
Its turn starts in the continuation of $AB$ beyond $B$ and ends along $BC$ .
Extending $BC$ backward instead makes the exterior turn for the reversed order, from $CB$ toward $BA$ , which is a right turn.
Keeping this choice only at $B$ mixes directions and does not describe the original boundary traversal.

Case B.
The positive measures alone cannot verify the direction record.
To repair it, restore the original left exterior choice at $B$ , or reverse the entire traversal and use right exterior choices at all three vertices.

Case A uses the three blue left turns. The red dashed alternative at B belongs to the reverse traversal.

Answer

Case A: at $A,B,C$ , the turns are $127^\circ,119^\circ,114^\circ$ , totaling $360^\circ$ . Case B: the claim is false; the sum remains $360^\circ$ , but the replacement reverses the turn direction at $B$ .

Key idea

A correct numerical exterior-angle total does not by itself check whether the chosen turns follow one traversal consistently.

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