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Sides, Angles, and Shortest Distances: 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 Read a triangle's angle order

Triangle UVW has angles U, V, and W measuring 39, 74, and 67 degrees, respectively. List its sides from shortest to longest, explaining each angle-side match.

The labels and markings show the givens.

Text description of this figure

Triangle UVW has given angles U, V, and W of 39, 74, and 67 degrees. No side order is marked.
1. Hint 1
  Match each angle with the side that does not touch its vertex.
2. Hint 2
  Put the angle measures in increasing order before ordering their opposite sides.
Answer

Shortest to longest: VW, UV, UW.

Full solution

Angle U is smallest, so the side opposite it, VW, is shortest.

Angle W is between U and V in measure, so its opposite side UV comes next.

Angle V is largest, so its opposite side UW is longest.

The comparisons are
$VW<UV.$
Then
$UV<UW.$

Answer

Shortest to longest: VW, UV, UW.

Key idea

Ranking angles ranks their opposite sides within the same triangle.
Problem 2 Read a triangle's side order

An existing triangle PQR has side lengths PQ of 11 units, PR of 17 units, and QR of 13 units. Order its angles from smallest to largest. Explain why a rotated drawing would give the same answer.

The labels and markings show the givens.

Text description of this figure

Triangle PQR has given lengths PQ of 11, PR of 17, and QR of 13 units. No angle order is marked.
1. Hint 1
  Identify the opposite angle for each given side.
2. Hint 2
  A rotation preserves the given lengths; the opposite-side correspondence also stays the same.
Answer

Smallest to largest: R, P, Q. A rotation does not change this order.

Full solution

Side PQ is shortest and faces angle R.

Side QR is next and faces angle P.

Side PR is longest and faces angle Q.

Thus
$m\angle R<m\angle P.$
Also
$m\angle P<m\angle Q.$
Rotation changes the drawing's orientation, not its lengths or angle measures.

Answer

Smallest to largest: R, P, Q. A rotation does not change this order.

Key idea

Opposite-side comparisons depend on the triangle, not its orientation.
Problem 3 Use an extended side

In triangle ABC, extend BC through C to D. The exterior angle ACD measures 131 degrees.

The remote angle at A is
$m\angle A=(4t+7)^\circ.$
The remote angle at B is
$m\angle B=(2t+10)^\circ.$
Find t and all three interior angles. Identify the longest and shortest sides.
1. Hint 1
  The exterior angle is the sum of the two remote interior angles.
2. Hint 2
  After finding t, use the straight angle at C and compare all three interior angles.
Answer

$\begin{aligned}t&=19.\\m\angle A&=83^\circ.\\m\angle B&=48^\circ.\\m\angle C&=49^\circ.\end{aligned}$
Longest side: BC. Shortest side: AC.

Full solution

The two remote angle measures add to the exterior angle:
$6t+17=131.$

Subtract 17, then divide by 6:
$t=19.$

Substitution gives
$m\angle A=4(19)+7.$
Hence
$m\angle A=83^\circ.$

For angle B,
$m\angle B=2(19)+10.$
Hence
$m\angle B=48^\circ.$

The angle at C is supplementary to the exterior angle:
$m\angle C=180^\circ-131^\circ.$
Hence
$m\angle C=49^\circ.$

Angle A is largest, so BC is longest.
Angle B is smallest, so AC is shortest.

The labels and markings show the givens.

Answer

$\begin{aligned}t&=19.\\m\angle A&=83^\circ.\\m\angle B&=48^\circ.\\m\angle C&=49^\circ.\end{aligned}$
Longest side: BC. Shortest side: AC.

Key idea

Find missing angles before translating their order into side order.
Problem 4 Equal angles with unequal expressions

An existing triangle LMN has equal angles at M and N.

The length of LM is
$LM=2x+5.$
The length of LN is
$LN=3x-2.$
Side MN has length 12 units. Find x and the lengths of LM and LN. Identify every largest angle.
1. Hint 1
  Equal angles have equal opposite sides.
2. Hint 2
  Once the expressions match, compare their common value with the third side.
Answer

$\begin{aligned}x&=7.\\LM&=19.\\LN&=19.\end{aligned}$
Angles M and N tie for largest.

Full solution

The equal angles at M and N face LN and LM, respectively, so
$2x+5=3x-2.$

Subtract 2x, then add 2:
$x=7.$

Evaluate each length separately:
$LM=2(7)+5.$
Thus
$LM=19.$

Likewise,
$LN=3(7)-2.$
Thus
$LN=19.$

Both equal sides are longer than MN, which is 12.
Their opposite angles M and N are equal and both larger than angle L.

The labels and markings show the givens.

Answer

$\begin{aligned}x&=7.\\LM&=19.\\LN&=19.\end{aligned}$
Angles M and N tie for largest.

Key idea

Equality cases belong in an angle-side comparison just as strict inequalities do.
Problem 5 A path to a straight bank

A straight river bank is modeled by a line. Camp P lies outside that line, and H is the perpendicular foot from P. The distance PH is 18 meters.

A path from P ends at a different point Q on the bank. Its length, in meters, is
$PQ=3x+6.$
Find the necessary restriction on x from comparing the path with PH. Explain why the boundary value is excluded.

Only the givens and incidence are marked.

Text description of this figure

P is outside a line. PH meets it at a marked right angle, and Q is another point on it. PQ is slanted.
1. Hint 1
  View the two segments as sides of a right triangle.
2. Hint 2
  The slanted segment is the hypotenuse and must be strictly longer than the perpendicular.
Answer

$x>4.$
The boundary value 4 is excluded.

Full solution

Triangle PHQ is right at H, so PQ is its hypotenuse.

The necessary length comparison is
$3x+6>18.$

Subtract 6 and divide by positive 3:
$x>4.$

At x equal to 4, PQ would measure 18 meters, tying PH.
A different endpoint Q gives a strictly longer segment, so that value is excluded.

Answer

$x>4.$
The boundary value 4 is excluded.

Key idea

Distance to a line uses the perpendicular, with strict inequality for every different endpoint.
Problem 6 Which angle can become largest?

Triangle ABC has AB shorter than AC, and angle B measures 70 degrees. Let angle C measure c degrees.

Find the possible range of c.

A largest: Determine when angle A is the unique largest angle.

A and B tie: Determine when angles A and B tie for largest.

B largest: Determine when angle B is the unique largest angle.

The labels and markings show the givens.

Text description of this figure

Triangle ABC has the given angle B of 70 degrees, angle C denoted c degrees, and the stated comparison AB less than AC. No range or largest-angle case is marked.
1. Hint 1
  Translate the side comparison into a comparison between angles B and C.
2. Hint 2
  Express angle A using the angle sum, then compare it with 70 degrees.
Answer

$0<c<70.$
Angle A is largest for
$0<c<40.$
Angles A and B tie for largest at
$c=40.$
Angle B is largest for
$40<c<70.$

Full solution

Because AC exceeds AB, its opposite angle B exceeds angle C.
Positivity then gives
$0<c<70.$

The angle sum gives
$m\angle A=(110-c)^\circ.$

Conversely, every c in this range gives positive angles of 70, c, and 110-c degrees summing to 180 degrees, so a triangle exists; B exceeds C, which gives AC greater than AB.

For A to exceed B, solve
$110-c>70.$
This gives
$c<40.$

A largest: When c lies strictly between 0 and 40, A exceeds B.
B already exceeds C, so A is largest.

A and B tie: At c equal to 40, both A and B measure 70 degrees.
They tie for largest.

B largest: When c lies strictly between 40 and 70, A is below 70 degrees.
B also exceeds C, so B is largest.

Answer

$0<c<70.$
Angle A is largest for
$0<c<40.$
Angles A and B tie for largest at
$c=40.$
Angle B is largest for
$40<c<70.$

Key idea

Partial order information can leave several cases, including an equality boundary.
Problem 7 An obtuse corner

Triangle ABC has angle A of 110 degrees. Explain why BC is longer than both other sides without finding their lengths.

Then justify the same conclusion whenever angle A is obtuse.

The labels and markings show the givens.

Text description of this figure

Triangle ABC has the given obtuse angle A of 110 degrees. The other angle measures and side order are not labeled.
1. Hint 1
  Compare the total of the two remaining angles with the obtuse angle.
2. Hint 2
  In the general case, an obtuse angle leaves less than 90 degrees for both other angles together.
Answer

BC is uniquely longest, both for the given triangle and whenever angle A is obtuse.

Full solution

For the given triangle, the other angles sum to
$m\angle B+m\angle C=70^\circ.$

Each is positive, so each is less than 70 degrees and therefore less than the 110-degree angle A.

The side opposite A is BC, which is longer than each side opposite a smaller angle.

In general, an obtuse angle is greater than 90 degrees.
The other two positive angles together total less than 90 degrees, so each is smaller than A.

The larger-angle theorem again makes BC uniquely longest.

Answer

BC is uniquely longest, both for the given triangle and whenever angle A is obtuse.

Key idea

The triangle angle sum identifies a largest angle even when the other two measures are unknown.
Problem 8 Compare across a shared side

Triangles ABC and ACD share side AC. B and D lie on opposite sides of line AC.

In triangle ABC, the given comparison is
$m\angle ABC>m\angle BAC.$
In triangle ACD, the given comparison is
$m\angle ACD>m\angle ADC.$
Determine whether AD or BC is longer. Justify every comparison you use.

The labels and markings show the givens.

Text description of this figure

Triangles ABC and ACD share AC, with B above AC and D below it. The legend states the given angle comparisons ABC greater than BAC and ACD greater than ADC. No derived side comparison is marked.
1. Hint 1
  First make a comparison within each triangle.
2. Hint 2
  The shared length AC can connect the two separate comparisons.
Answer

$AD>BC.$

Full solution

In triangle ABC, the larger angle ABC faces AC.
The smaller angle BAC faces BC, so
$AC>BC.$

In triangle ACD, the larger angle ACD faces AD.
The smaller angle ADC faces AC, so
$AD>AC.$

A length larger than AC is also larger than every length below AC.
Therefore
$AD>BC.$

Answer

$AD>BC.$

Key idea

Comparisons across triangles need a justified link such as a shared side.
Problem 9 Compare an isosceles triangle's angles

A triangle has exactly two equal sides. Its third side is shorter than either equal side. Can the triangle have a unique largest angle? Explain the complete angle order.
1. Hint 1
  Match the equal sides to their opposite angles.
2. Hint 2
  Compare each equal side with the shorter third side before deciding whether any largest angle is unique.
Answer

No. The two angles opposite the equal sides tie for largest; the angle opposite the third side is uniquely smallest.

Full solution

Equal sides face equal angles, so the angles opposite the two equal sides have the same measure.

Each equal side is longer than the third side.
Each of their opposite angles is therefore larger than the angle opposite the third side.

The two equal angles tie for largest, and the remaining angle is uniquely smallest.

The equal-largest angle markings are conclusions shown only with the solution.

Answer

No. The two angles opposite the equal sides tie for largest; the angle opposite the third side is uniquely smallest.

Key idea

A strict comparison with a third side can coexist with equality between the other two sides.
Problem 10 Three segments, one line

H, Q, and R lie in that order on a line. Point P is outside the line, and PH is perpendicular to it.

The given angle from PH to PQ is
$m\angle HPQ=28^\circ.$
The given angle from PH to PR is
$m\angle HPR=53^\circ.$
Find angles PQR and QPR. Order PH, PQ, and PR from shortest to longest, with geometric reasons.
1. Hint 1
  Identify angle PQR as an exterior angle of right triangle PHQ.
2. Hint 2
  Subtract the two given angles at P to find angle QPR.
3. Hint 3
  Compare PQ with PH in the right triangle, and compare PR with PQ in triangle PQR.
Answer

$\begin{aligned}m\angle PQR&=118^\circ.\\m\angle QPR&=25^\circ.\end{aligned}$
Shortest to longest: PH, PQ, PR.

Full solution

Because Q is between H and R, ray QR extends side QH through Q.
Angle PQR is exterior to triangle PHQ.

Its remote angles are the right angle at H and angle HPQ, so
$m\angle PQR=90^\circ+28^\circ.$
Hence
$m\angle PQR=118^\circ.$

Ray PQ lies inside angle HPR because the perpendicular foot and the two endpoints occur in the stated order.
Subtract the nested angles at P:
$m\angle QPR=53^\circ-28^\circ.$
Hence
$m\angle QPR=25^\circ.$

Angle PQR is obtuse, so it is the largest angle of triangle PQR.
Its opposite side PR is longer than PQ.

In right triangle PHQ, PQ is the hypotenuse and exceeds PH.

Therefore the order is PH, PQ, PR.

The angles of triangle PQR sum to 180 degrees, so
$m\angle PRQ=180^\circ-118^\circ-25^\circ.$

Hence
$m\angle PRQ=37^\circ.$

Numerical angle labels are conclusions, revealed only with the solution.

Answer

$\begin{aligned}m\angle PQR&=118^\circ.\\m\angle QPR&=25^\circ.\end{aligned}$
Shortest to longest: PH, PQ, PR.

Key idea

Choose the triangle that contains both lengths being compared.

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