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Angles in a Triangle: 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 Two classification records

Classify each nondegenerate triangle using only the supplied information.

Case A. A triangle has side lengths $12,12,12$ centimeters. Give every applicable side classification.

Case B. A triangle has interior angle measures $41^\circ,52^\circ,87^\circ$ . Give its angle classification.
1. Hint 1
  For Case A, compare the equal side lengths with both the isosceles and equilateral definitions.
2. Hint 2
  For Case B, compare each angle measure with 90 degrees.
Answer

Case A: equilateral and isosceles. Case B: acute.

Full solution

Case A.
All three sides have equal lengths, so they are congruent and the triangle is equilateral.
It also has at least two congruent sides, so it is isosceles under the inclusive definition.

Case B.
All three angle measures are positive and less than $90^\circ$ , so the triangle is acute.
They also have the required total, since
$41+52+87=180.$

Answer

Case A: equilateral and isosceles. Case B: acute.

Key idea

Choose a side classification from side information and an angle classification from angle information.
Problem 2 The third corner

In triangle $RST$ , $m\angle SRT=67^\circ$ and $m\angle RST=78^\circ$ . Find $m\angle RTS$ and classify the triangle by angles.

The labels record the stated givens.

Text description of this figure

Triangle RST has angle 67 degrees at R and 78 degrees at S; the angle at T is unknown.
1. Hint 1
  All three interior angles belong to the same triangle.
2. Hint 2
  Subtract the two known measures from the triangle total, then compare all three measures with 90 degrees.
Answer

$m\angle RTS=35^\circ$ ; acute.

Full solution

The triangle angle-sum theorem gives a total of 180 degrees for the three interior angles.
Subtract the two given measures.

The remaining measure in degrees is
$180-67-78=35.$

Thus
$m\angle RTS=35^\circ.$
Each of $67^\circ,78^\circ,35^\circ$ is acute, so the triangle is acute.
The measures are positive and total $180^\circ$ .

Answer

$m\angle RTS=35^\circ$ ; acute.

Key idea

The third interior angle is the positive part left after two known measures are subtracted from 180 degrees.
Problem 3 Complementary expressions

Triangle $LMN$ is right at $L$ . The other measures are $m\angle LMN=(3r+2)^\circ$ and $m\angle LNM=(2r+8)^\circ$ . Find $r$ and both acute angle measures.

The square records the given right angle.

Text description of this figure

Triangle LMN is drawn with a right-angle square at L. The acute angles have no numeric labels.
1. Hint 1
  Account for the given right angle before using the two expressions.
2. Hint 2
  The two unknown measures total 90 degrees. Write one equation using both expressions.
3. Hint 3
  Combine like terms, solve for r, and substitute into each original expression.
Answer

$r=16$ ; $m\angle LMN=50^\circ$ and $m\angle LNM=40^\circ$ .

Full solution

The right angle uses $90^\circ$ of the $180^\circ$ triangle total, so the two remaining measures are complementary.

$(3r+2)+(2r+8)=90.$

Combining terms gives $5r+10=90$ .
Subtracting $10$ gives $5r=80$ , and dividing by $5$ gives $r=16$ .

At $M$ , substitute to obtain $3(16)+2=50$ , so
$m\angle LMN=50^\circ.$

At $N$ , substitute to obtain $2(16)+8=40$ , so
$m\angle LNM=40^\circ.$
Their sum is $90^\circ$ , as required.

Answer

$r=16$ ; $m\angle LMN=50^\circ$ and $m\angle LNM=40^\circ$ .

Key idea

A given right angle makes the other two interior angle measures sum to 90 degrees.
Problem 4 Match the unused vertices

Triangle $GHI$ has $54^\circ$ at $G$ and $71^\circ$ at $H$ . Triangle $JKL$ has $71^\circ$ at $J$ and $54^\circ$ at $K$ . Find the remaining angle in each triangle. Then name all three angle correspondences justified by the data and the third-angle theorem.
1. Hint 1
  Match equal given measures without using the alphabetical order of the letters.
2. Hint 2
  Each triangle has the same two known measures, so each subtracts the same total from 180 degrees.
Answer

$m\angle GIH=m\angle JLK=55^\circ$ ; $\angle G\cong\angle K$ , $\angle H\cong\angle J$ , and $\angle I\cong\angle L$ .

Full solution

The givens pair $\angle G$ with $\angle K$ , since both measure $54^\circ$ .
They pair $\angle H$ with $\angle J$ , since both measure $71^\circ$ .

The remaining measure in each triangle is
$180^\circ-54^\circ-71^\circ=55^\circ.$

Thus $m\angle GIH=55^\circ$ and
$m\angle JLK=55^\circ.$
Equal measures give $\angle I\cong\angle L$ , as the third-angle theorem predicts.
No corresponding side-length equality is needed or concluded.

Answer

$m\angle GIH=m\angle JLK=55^\circ$ ; $\angle G\cong\angle K$ , $\angle H\cong\angle J$ , and $\angle I\cong\angle L$ .

Key idea

Match the supplied angle pairs before identifying which two vertices remain.
Problem 5 Prove the sum from another vertex

Let $ABC$ be a nondegenerate Euclidean triangle with $\angle ACB$ obtuse. Orient your paper sketch with $B$ to the left of $C$ on a horizontal base and $A$ above and to the right of $C$ , as shown. Your proof must apply to every triangle with this stated arrangement, without measuring the drawing.

Draw an auxiliary line through $A$ parallel to $BC$ . Choose points $D,E$ on it with $D$ left of $A$ and $E$ right of $A$ . Mark the two alternate interior angle pairs and write a paragraph proof that the three interior angle measures of $ABC$ sum to $180^\circ$ . Name the accepted foundation that allows the new parallel. Use straight-angle addition rather than the triangle angle-sum theorem as a premise.

Add the requested construction on paper.

Text description of this figure

Triangle ABC has B to the left of C on a horizontal base, and A above and to the right of C. Angle ACB is given to be obtuse. No auxiliary line, angle equality, or derived measure is drawn.
1. Hint 1
  Use the three noncollinear vertices to check that A is outside line BC before choosing a parallel.
2. Hint 2
  Treat AB and AC as separate transversals. In the stated orientation, follow the angles below the new line from its leftward ray to its rightward ray.
3. Hint 3
  Write angle addition for the straight angle at A, then replace the two outside pieces with their congruent triangle angles.
Answer

Accepted foundation: a unique parallel through a point outside a line in the same plane. Alternate interior pairs: $\angle DAB\cong\angle ABC$ and $\angle CAE\cong\angle ACB$ . The interior measures total $180^\circ$ .

Full solution

Because $A,B,C$ are noncollinear, $A$ lies outside line $BC$ .
The accepted parallel foundation supplies a line through $A$ parallel to $BC$ .
Choose $D$ and $E$ on its opposite rays in the stated left-to-right order $D,A,E$ .

With $AB$ as a transversal of the parallel lines $DE$ and $BC$ , $\angle DAB$ and $\angle ABC$ are alternate interior angles.
Therefore $\angle DAB\cong\angle ABC$ .

With $AC$ as the transversal, $\angle CAE$ and $\angle ACB$ are alternate interior angles.
Therefore $\angle CAE\cong\angle ACB$ .
This pair is obtuse in the chosen arrangement; alternate interior congruence applies to obtuse angles as well.

Both $B$ and $C$ are below $DE$ .
Since $B$ is left of $C$ and $A$ is above and right of $C$ , the rays occur below $DE$ in the order $AD,AB,AC,AE$ .
Thus $\angle DAB$ , $\angle BAC$ , and $\angle CAE$ are adjacent parts of the straight angle $\angle DAE$ .

Write $p=m\angle DAB$ , $q=m\angle BAC$ , and
$r=m\angle CAE.$
Straight-angle addition gives
$p+q+r=180^\circ.$

The two alternate interior congruences give $p=m\angle ABC$ and
$r=m\angle ACB.$
Substituting those equal measures shows that the three original interior angle measures of $ABC$ sum to $180^\circ$ .
No triangle-sum or right-angle restriction was assumed in this argument.

Matching arcs show the two justified alternate interior pairs; p, q, and r fill the straight angle at A.

Answer

Accepted foundation: a unique parallel through a point outside a line in the same plane. Alternate interior pairs: $\angle DAB\cong\angle ABC$ and $\angle CAE\cong\angle ACB$ . The interior measures total $180^\circ$ .

Key idea

An auxiliary parallel can place matching copies of two triangle angles beside the third at a different vertex.
Problem 6 Two right triangles inside one

Triangle $ABC$ is right at $B$ . Point $D$ lies strictly between $A$ and $C$ , and $BD\perp AC$ . Given $m\angle BAC=32^\circ$ , find $m\angle ACB$ , $m\angle ABD$ , and $m\angle DBC$ . Justify the transfer of any angle from the large triangle to a smaller one.

The two squares record the given perpendicular relationships.

Text description of this figure

Triangle ABC is right at B. D lies inside AC and BD is perpendicular to AC, with right-angle squares at B and D. The angle at A is labeled 32 degrees. No other measure is displayed.
1. Hint 1
  Identify the right angle in each triangle you use.
2. Hint 2
  First use the whole triangle to find its angle at C. In triangle ABD, the angle at A is the original 32-degree angle.
3. Hint 3
  Use the right angle at D in each smaller triangle; no side-length relationship is needed.
Answer

$m\angle ACB=58^\circ$ , $m\angle ABD=58^\circ$ , and $m\angle DBC=32^\circ$ .

Full solution

In the whole right triangle, the acute angles are complementary, so
$m\angle ACB=90^\circ-32^\circ=58^\circ.$

Since $D$ lies on $AC$ , rays $AD$ and $AC$ agree.
Therefore the angle at $A$ in triangle $ABD$ is the given $32^\circ$ angle.
The perpendicularity gives a right angle at $D$ .

The triangle sum in $ABD$ gives the remaining measure in degrees:
$180-90-32=58.$
Thus
$m\angle ABD=58^\circ.$

Similarly, rays $CD$ and $CA$ agree, so the angle at $C$ in triangle $BDC$ is $58^\circ$ .
Its angle at $D$ is right.
The remaining measure in degrees is
$180-90-58=32.$
Hence
$m\angle DBC=32^\circ.$

As a local check, the adjacent parts of the right angle at $B$ total
$58^\circ+32^\circ=90^\circ.$

Answer

$m\angle ACB=58^\circ$ , $m\angle ABD=58^\circ$ , and $m\angle DBC=32^\circ$ .

Key idea

A smaller triangle inherits a larger triangle angle when its two defining rays agree with the original rays.
Problem 7 A shared acute measure

Triangle $PQR$ has $118^\circ$ at $P$ and $22^\circ$ at $Q$ . Triangle $UVW$ has $\angle U\cong\angle R$ and $35^\circ$ at $V$ . Find $m\angle R$ , $m\angle W$ , and the angle classification of each triangle. Explain why triangle UVW cannot also have a right angle, using the angle sum and the positivity of interior angles.
1. Hint 1
  Find the remaining angle of the first triangle before transferring a measure.
2. Hint 2
  Congruent angles have equal measures; that gives one measure in the second triangle.
3. Hint 3
  Use a separate angle-sum equation in the second triangle.
4. Hint 4
  Compare the obtuse measure with a hypothetical 90-degree angle; the third interior angle would still have to be positive.
Answer

$m\angle R=40^\circ$ and $m\angle W=105^\circ$ ; both triangles are obtuse, and $UVW$ cannot also have a right angle.

Full solution

The first triangle has
$m\angle R=180^\circ-118^\circ-22^\circ=40^\circ.$

The given congruence gives
$m\angle U=40^\circ.$
In the second triangle,
$m\angle W=180^\circ-40^\circ-35^\circ=105^\circ.$

Triangle $PQR$ is obtuse because of its $118^\circ$ angle.
Triangle $UVW$ is obtuse because of its $105^\circ$ angle.

If another interior angle of $UVW$ were right, those two measures would already total
$105^\circ+90^\circ=195^\circ.$
The third interior angle must be positive, so the total would be greater still and could not equal $180^\circ$ .
Therefore $UVW$ cannot also have a right angle.

Answer

$m\angle R=40^\circ$ and $m\angle W=105^\circ$ ; both triangles are obtuse, and $UVW$ cannot also have a right angle.

Key idea

Any obtuse angle measures more than 90 degrees, so adding a right angle already exceeds the triangle total before the positive third angle is included.
Problem 8 A three-point arrangement

A claim says that joining any three distinct points always forms a triangle, so the triangle angle-sum theorem always applies. Consider distinct collinear points $R,S,T$ , with $S$ between $R$ and $T$ . Decide whether the theorem applies to this arrangement, and identify the condition that the claim omitted.
1. Hint 1
  Check the definition of a triangle before adding angle measures.
2. Hint 2
  Compare the positions of these three points with the hypothesis required for the triangle theorem.
Answer

The theorem does not apply; a triangle requires three noncollinear vertices.

Full solution

The points $R,S,T$ all lie on one line.
Joining the pairs gives overlapping straight segments rather than three sides enclosing a nondegenerate triangle.

The claim omitted noncollinearity.
Distinctness alone does not make three points triangle vertices, so the triangle angle-sum theorem does not apply to this arrangement.

The joining segments collapse onto one straight segment.

Answer

The theorem does not apply; a triangle requires three noncollinear vertices.

Key idea

Check that an object satisfies a theorem’s hypotheses before applying its conclusion.
Problem 9 Combine a bisector with a linear pair

In triangle $ABC$ , $D$ is strictly between $A$ and $B$ , and $CD$ bisects $\angle ACB$ . Given $m\angle CAB=48^\circ$ and $m\angle CDB=96^\circ$ , find $m\angle ABC$ and $m\angle ACB$ . Justify each intermediate angle.

The labels record the stated givens.

Text description of this figure

Triangle ABC has D between A and B, joined to C. The angle at A is 48 degrees and angle CDB is 96 degrees. Separate arcs with matching single tick marks record that CD bisects the angle at C; the two top measures are not displayed.
1. Hint 1
  The two angles next to each other at D lie along the same straight base.
2. Hint 2
  Find angle ADC first, then use triangle ACD to determine its angle at C.
3. Hint 3
  The bisector supplies the matching angle at C in triangle CDB.
Answer

$m\angle ABC=36^\circ$ and $m\angle ACB=96^\circ$ .

Full solution

Rays $DA,DB$ are opposite, so $\angle ADC$ and $\angle CDB$ form a linear pair.
Thus
$m\angle ADC=180^\circ-96^\circ=84^\circ.$

Since rays $AD,AB$ agree, triangle $ACD$ has $48^\circ$ at $A$ .
Its angle sum gives the remaining measure in degrees:
$180-48-84=48.$
Thus
$m\angle ACD=48^\circ.$

The bisector gives
$m\angle DCB=48^\circ.$
Angle addition gives
$m\angle ACB=48^\circ+48^\circ=96^\circ.$

In triangle $CDB$ , the remaining measure in degrees is
$180-96-48=36.$
Therefore
$m\angle CBD=36^\circ.$

Rays $BD,BA$ agree, so
$m\angle ABC=36^\circ.$
The whole triangle checks because
$48+36+96=180.$

Answer

$m\angle ABC=36^\circ$ and $m\angle ACB=96^\circ$ .

Key idea

A linear pair can supply one triangle angle, and a bisector can transfer the next angle into an adjacent triangle.
Problem 10 Choose a route through the diagram

In triangle $ABC$ , $D$ lies strictly inside side $AB$ and $E$ lies strictly inside side $AC$ , with $DE\parallel BC$ . Point $F$ lies strictly between $D$ and $E$ , and $AF$ bisects $\angle BAC$ . The given measures are $m\angle ABC=52^\circ$ and $m\angle ACB=68^\circ$ .

Find $m\angle AFD$ and $m\angle AFE$ . Begin with a short plan naming the angles you will determine before the targets. Then justify every step of your chase.

Incidence, parallelism, and the stated base measures are given.

Text description of this figure

In triangle ABC, D lies inside AB and E lies inside AC. Matching chevrons record that dashed segment DE is parallel to blue base BC. F lies between D and E, and separate arcs with matching single ticks mark the two halves made by AF. Only the given angles, 52 degrees at B and 68 degrees at C, are displayed.
1. Hint 1
  Choose a triangle containing one target angle and ask which of its other angles can be found.
2. Hint 2
  Use the large triangle and the bisector to find the angle at A in triangle ADF.
3. Hint 3
  Use DE parallel to BC to transfer the angle at B to D. After finding one target, inspect the straight line through D,F,E.
Answer

$m\angle AFD=98^\circ$ and $m\angle AFE=82^\circ$ .

Full solution

One plan is to find the whole angle at A, halve it, transfer the base angle to D using the parallels, and use triangle ADF.
The other target then follows from the line at F.

The whole triangle has
$m\angle BAC=180^\circ-52^\circ-68^\circ=60^\circ.$

Since $AF$ bisects that angle and rays $AD,AB$ agree,
$m\angle DAF=30^\circ.$

For parallel lines $DE,BC$ with transversal $AB$ , $\angle ADF$ and $\angle ABC$ are corresponding angles: rays $DF,DE$ agree because $F$ lies between $D$ and $E$ .
Therefore
$m\angle ADF=52^\circ.$

Triangle $ADF$ now gives
$m\angle AFD=180^\circ-30^\circ-52^\circ=98^\circ.$

Rays $FD,FE$ are opposite, so $\angle AFD$ and $\angle AFE$ form a linear pair.
Thus
$m\angle AFE=180^\circ-98^\circ=82^\circ.$

An alternative check uses triangle $AFE$ : its angles at $A,E$ are $30^\circ,68^\circ$ by the bisector and corresponding angles.
Their sum leaves $82^\circ$ at $F$ , agreeing with the chase.

These measures appear only after the written justifications.

Answer

$m\angle AFD=98^\circ$ and $m\angle AFE=82^\circ$ .

Key idea

Plan backward from a target triangle, then work forward using only justified angles.

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