MAT8 Q2W7D2: Triangle Inequality & Third-Side Range

By the end of the lesson, you will be able to:

1. Compare sides and angles within a triangle: Given any triangle with side lengths labeled, correctly order its interior angles from least to greatest (and vice versa from angle measures to side lengths) in at least 8 items with ≥ 85% accuracy using the fact that the larger side lies opposite the larger angle.
2. Apply the Hinge Theorem (SAS Inequality): For two triangles with two pairs of corresponding sides congruent, determine which triangle has the longer included-opposite side (or the larger included angle) using the Hinge Theorem and its converse in at least 6 problems with complete justifications.
3. Solve applied inequality problems: Model and solve real-world scenarios that require side–angle comparisons or the SAS inequality; state the conclusion with the correct inequality symbol $>$ or $<$ in at least 4 application problems with clear reasoning.

• Included angle – the angle formed by two given sides.
• Hinge Theorem (SAS Inequality) – If two sides of $\Delta ABC$ are congruent to two sides of $\Delta DEF$ and the included angle in the first is larger, then the side opposite that angle is longer: $$(\mathrm{AB}=\mathrm{DE},\mathrm{AC}=\mathrm{DF},\angle A>\angle D)⟹\mathrm{BC}>\mathrm{EF}$$
• Converse of the Hinge Theorem – With two pairs of corresponding sides congruent, if one triangle’s third side is longer, then its included angle is larger.
• Side–angle ordering in a single triangle – Larger side ↔ larger opposite angle.
• Inequality symbols – $>$, $<$, and correct use in geometric comparisons.

A) Match side–angle ordering (within one triangle)
In $\Delta PQR$, suppose $\mathrm{PQ}<\mathrm{PR}<\mathrm{QR}$. Order the angles from smallest to largest.

Show Answer

Longer sides face larger angles. Since $\mathrm{QR}$ is longest, $\angle P$ is largest. Since $\mathrm{PQ}$ is shortest, $\angle R$ is smallest. Order: $\angle R<\angle Q<\angle P$.

B) Identify the included angle
In $\Delta ABC$, for sides $\mathrm{AB}$ and $\mathrm{AC}$, name the included angle.

Show Answer

The included angle is the angle formed by the two sides at their common endpoint: $\angle A$.

C) Day 1 connection (third-side range refresher)
Two sides measure $8$ cm and $13$ cm. State the open interval for the third side $x$.

Show Answer

Use difference–sum bounds: $|13-8|<x<13+8$ → $5<x<21$.

D) Quick compare using sides ↔ angles
In $\Delta JKL$, suppose $\angle J>\angle K$. Which side is longer: $\mathrm{KL}$ or $\mathrm{JL}$?

Show Answer

Larger angle faces the longer opposite side. $\angle J>\angle K$ implies $\mathrm{KL}>\mathrm{JL}$.

0) Orientation and learning map

Yesterday you learned to decide whether three lengths can form a triangle and how the third side is bounded by the sum and the positive difference of the other two. Today you will explore how side lengths and angles compare inside a triangle and how to compare two different triangles when two pairs of sides match but the included angle differs. By the end, you will be able to order angles when the sides are known, and order sides when the angles are known, use the Hinge Theorem to compare non-included sides in two triangles with two corresponding sides congruent, use the converse of the Hinge Theorem to compare included angles when two corresponding sides and the opposite sides are known, and apply these ideas to practical settings like linkages, braces, navigation corners, and sports spacing. Throughout this Explore, you will see guiding questions and checkpoints. Try them before opening the answers. After each checkpoint you will find a mini-summary to cement the idea.

1) Warm-up intuition: which side faces the larger angle?

Imagine opening and closing a paper triangle where two sides are fixed. As you increase the angle at a vertex, the side across from that angle must adjust. Your experience from Day 1 suggests the opposite side gets longer as the included angle gets wider.

1.1 Side–angle pairing inside one triangle

Label a triangle with side lengths $a$, $b$, $c$ opposite angles $\angle A$, $\angle B$, $\angle C$ respectively. The basic comparison rule in one triangle is:

$$a>b\iff \angle A>\angle B,b>c\iff \angle B>\angle C,c>a\iff \angle C>\angle A$$

This can be summarized as: larger side ↔ larger opposite angle. The ordering among the three sides matches the ordering among their opposite angles.

Guiding question S1

In $\Delta PQR$, suppose $\mathrm{PQ}>\mathrm{PR}>\mathrm{QR}$. Order the angles from largest to smallest.

Show Answer

The longest side faces the largest angle. The order of sides is $\mathrm{PQ}>\mathrm{PR}>\mathrm{QR}$, so angles opposite them order as $\angle R>\angle Q>\angle P$.

Guiding question S2

In $\Delta ABC$, if $\angle B>\angle C$, which side is longer: $\mathrm{AC}$ or $\mathrm{AB}$?

Show Answer

$\angle B$ is larger, so the side opposite it, $\mathrm{AC}$, is longer than the side opposite $\angle C$ which is $\mathrm{AB}$. Thus $\mathrm{AC}>\mathrm{AB}$.

Mini-summary 1 - inside a single triangle the ranking of sides matches the ranking of opposite angles. If you know any two, you immediately know how the third compares.

2) From intuition to a precise comparative theorem

You now formalize the observation with a clean statement. Consider two triangles that share two corresponding side lengths, and you vary the included angle. The question is: which triangle has the longer third side?

2.1 The Hinge Theorem (SAS Inequality)

Statement: If two sides of one triangle are respectively congruent to two sides of another triangle, and the included angle in the first triangle is greater than the included angle in the second, then the side opposite the included angle in the first triangle is longer than the corresponding side in the second.

Using letters, suppose triangles are $\Delta ABC$ and $\Delta DEF$, with $\mathrm{AB}=\mathrm{DE}$, $\mathrm{AC}=\mathrm{DF}$, and we compare the included angles $\angle A$ and $\angle D$. Then:

$$(\mathrm{AB}=\mathrm{DE},\mathrm{AC}=\mathrm{DF},\angle A>\angle D)⟹\mathrm{BC}>\mathrm{EF}$$

2.2 Converse of the Hinge Theorem

Statement: If two sides of one triangle are respectively congruent to two sides of another triangle, and the third side in the first triangle is longer than the third side in the second, then the included angle in the first is greater than the included angle in the second:

$$(\mathrm{AB}=\mathrm{DE},\mathrm{AC}=\mathrm{DF},\mathrm{BC}>\mathrm{EF})⟹\angle A>\angle D$$

Guiding question H1

Two triangles share two side lengths: $\mathrm{AB}=\mathrm{DE}=10$ and $\mathrm{AC}=\mathrm{DF}=7$. If $\angle A=68°$ and $\angle D=55°$, which is longer: $\mathrm{BC}$ or $\mathrm{EF}$?

Show Answer

The included angles are compared with two equal pairs of sides. Since $\angle A>\angle D$, the side opposite the larger angle in that triangle is longer. Therefore $\mathrm{BC}>\mathrm{EF}$.

Guiding question H2

With the same side pairs equal as above, if you learn instead that $\mathrm{BC}>\mathrm{EF}$, what can you conclude about $\angle A$ and $\angle D$?

Show Answer

Use the converse. If the third side in one triangle is longer, then the included angle in that triangle is larger. Hence $\angle A>\angle D$.

Mini-summary 2 - Hinge Theorem: two matching sides plus a larger included angle imply a longer opposite side. Converse: two matching sides plus a longer opposite side imply a larger included angle.

3) Within one triangle: ordering sides and angles efficiently

The hinge picture tells you how one side changes as the included angle changes when two sides are fixed. Within one triangle you compare parts of the same one. The relationship is consistent:

$$a>b\iff \angle A>\angle B.$$

Worked Example 1 - Order angles given sides

In $\Delta MNP$, the sides are $\mathrm{MN}=8$, $\mathrm{NP}=11$, $\mathrm{PM}=9$. Order the angles from smallest to largest.

Show Answer

The smallest side is $\mathrm{MN}=8$ so the smallest angle is opposite it: $\angle P$. Next is $\mathrm{PM}=9$ so $\angle N$ is the middle. Largest side $\mathrm{NP}=11$ implies largest angle $\angle M$. Order: $\angle P<\angle N<\angle M$.

Worked Example 2 - Order sides given angles

In $\Delta RST$, suppose $\angle R=32°$, $\angle S=71°$, $\angle T=77°$. Order the sides from longest to shortest.

Show Answer

Largest angle is $\angle T$, so longest side is $\mathrm{RS}$. Next is $\angle S$ so second longest is $\mathrm{RT}$. Smallest angle is $\angle R$ so shortest side is $\mathrm{ST}$. Order: $\mathrm{RS}>\mathrm{RT}>\mathrm{ST}$.

Checkpoint C1 - Try before peeking

1. In $\Delta ABC$, sides are $\mathrm{AB}=15$, $\mathrm{BC}=9$, $\mathrm{CA}=12$. Order the angles from largest to smallest.
2. In $\Delta UVW$, angles are $\angle U=40°$, $\angle V=49°$, $\angle W=91°$. Order the sides from shortest to longest.

Show Answer

1) Largest side is $\mathrm{AB}=15$ so largest angle is $\angle C$. Next is $\mathrm{CA}=12$ so next angle is $\angle B$. Smallest is $\mathrm{BC}=9$ so smallest angle is $\angle A$. Order: $\angle C>\angle B>\angle A$.

2) Smallest angle is $\angle U$ so shortest side is $\mathrm{VW}$. Next is $\angle V$ so middle side is $\mathrm{UW}$. Largest angle is $\angle W$ so longest side is $\mathrm{UV}$. Order: $\mathrm{VW}<\mathrm{UW}<\mathrm{UV}$.

Mini-summary 3 - the side–angle order rule is a fast mental tool. Always pair the largest with the largest opposite.

4) Comparing two triangles with a hinge

Consider you have two metal arms of the same lengths pinned at one end. If you open one at a wider angle, its tip-to-tip distance is longer. That is the comparison captured by the Hinge Theorem.

Worked Example 3 - Which base is longer?

Triangles $\Delta ABC$ and $\Delta DEF$ satisfy $\mathrm{AB}=\mathrm{DE}=10$, $\mathrm{AC}=\mathrm{DF}=8$. If $\angle A=52°$ and $\angle D=61°$, compare $\mathrm{BC}$ and $\mathrm{EF}$.

Show Answer

Two pairs of corresponding sides match. The included angles differ. Because $\angle D>\angle A$, the opposite side in triangle DEF is longer. Hence $\mathrm{EF}>\mathrm{BC}$.

Worked Example 4 - Use the converse to compare angles

With the same side pairs equal, suppose you measure $\mathrm{BC}=12.4$ and $\mathrm{EF}=11.9$. Which included angle is larger?

Show Answer

Third side in ABC is longer. With two corresponding sides equal, the triangle that has the longer third side has the larger included angle. Therefore $\angle A>\angle D$.

Guiding question H3

Two different frames use arms of length 30 cm and 18 cm. Frame X opens to $70°$, frame Y opens to $50°$. Which frame creates the longer cross-bar distance between free ends?

Show Answer

Same two side lengths with a larger included angle produce a longer opposite side. Frame X has the larger angle, so its cross-bar distance is longer.

Mini-summary 4 - same two sides plus larger included angle implies a longer opposite side. Same two sides plus longer opposite side implies a larger included angle.

5) Real-world applications

5.1 Linkages and adjustable braces

A shelf bracket may use two fixed-length bars anchored at a hinge. The diagonal spacer between their free ends changes length as the included angle changes. The Hinge Theorem predicts which setting yields a longer spacer.

Scenario L1 - Choosing a spacer
Two hinged bars are 42 cm and 27 cm. At setting A the angle is $40°$; at setting B the angle is $65°$. Which setting needs the longer spacer?

Show Answer

Same side lengths, larger included angle at B. So the spacer at B is longer.

Guiding question L2

How would you argue that the spacer at a smaller angle cannot be longer than at a larger angle if the two side lengths are fixed?

Show Answer

If the smaller angle produced a longer opposite side, it would contradict the Hinge Theorem which formalizes the monotonic relationship between included angle and opposite side when the adjacent sides are fixed. Intuitively, opening the hinge pulls the free ends apart.

5.2 Robotics and machine reach

A two-link planar robot arm has link lengths $L\_1$ and $L\_2$. If you fix the links and vary the elbow angle, the reach from shoulder to gripper is the side opposite that elbow angle. Larger included angle yields longer reach until the arm straightens.

Scenario R1 - Safety buffer
Links are $L\_1=35\mathrm{cm}$, $L\_2=20\mathrm{cm}$. Configuration X uses an elbow angle of $40°$; configuration Y uses $110°$. Which configuration risks colliding with a wall that is directly in front of the base?

Show Answer

Two matching links. The larger included angle at Y gives a longer straight-line reach, so Y risks collision first.

Guiding question R2

If the end-effector must never be closer than 16 cm to the base, and the links are 15 cm and 12 cm, is that constraint always satisfied for all elbow angles?

Show Answer

The reach range satisfies $|15-12|<r<27$, so $3<r<27$. The robot can reach as close as just above 3 cm, which violates the 16 cm constraint unless motion is limited. So the constraint is not automatically satisfied for all elbow angles.

5.3 Navigation corners and base distance

Two hikers start at the same point, walk side 1, turn by a certain angle, then walk side 2. The straight-line distance between start and end depends on the included turn angle. A larger turn angle yields a longer base distance between start and end - consistent with the Hinge Theorem applied to side lengths fixed.

Scenario N1 - Map triangle
Legs are 1.8 km and 1.2 km. Route A turns $50°$. Route B turns $100°$. Which route creates a longer direct start-to-end distance?

Show Answer

Same two sides with a larger included angle at B, so the base is longer in Route B.

5.4 Sports spacing and passing lanes

Three players form triangle vertices. If two distances are coached to remain fixed, then opening the included angle grows the third distance. That can be used to widen passing lanes intentionally.

Scenario S1 - Triangle drill
Two players keep 10 m and 14 m from a pivot. Drill A uses a 30° angle at the pivot, Drill B uses 80°. Which drill opens a larger lane opposite the pivot?

Show Answer

Drill B - larger included angle gives longer opposite side.

Mini-summary 5 - in machines, navigation, or team spacing, two fixed links and a larger included angle yield a longer opposite side. When a third side is longer under the same two sides, the included angle must be larger.

6) Algebraic thinking with comparisons and ranges

While the Hinge Theorem compares lengths qualitatively, you often also need quantitative reasoning. You can use comparisons like greater than or less than with MathML to keep statements precise.

Within one triangle:

$$\angle A>\angle B⟹a>b,b>c⟹\angle B>\angle C$$

Between two triangles with corresponding sides:

$$(\mathrm{AB}=\mathrm{DE},\mathrm{AC}=\mathrm{DF},\angle A>\angle D)⟹\mathrm{BC}>\mathrm{EF}$$

Worked Example 5 - Decide which triangle has the larger included angle

Two frames use struts of equal lengths $20$ and $12$. In one frame the third side measures 24 cm. In the other it measures 21 cm. Which frame has the larger included angle?

Show Answer

With two pairs of corresponding sides equal, compare the third sides. The frame with the longer third side has the larger included angle by the converse. The 24 cm frame has the larger included angle.

Worked Example 6 - Within-triangle side ordering from angle data

In $\Delta XYZ$, if $\angle X=44°$, $\angle Y=44°$, and $\angle Z=92°$, order the sides from shortest to longest and state if the triangle is isosceles.

Show Answer

Equal angles at X and Y imply equal opposite sides $\mathrm{YZ}$ and $\mathrm{XZ}$. Largest angle at Z makes the side opposite, $\mathrm{XY}$, the longest. Thus $\mathrm{YZ}=\mathrm{XZ}<\mathrm{XY}$. The triangle is isosceles.

Checkpoint A1 - Short tasks

1. With $\mathrm{AB}=\mathrm{DE}$, $\mathrm{AC}=\mathrm{DF}$, and $\mathrm{BC}>\mathrm{EF}$, compare $\angle A$ and $\angle D$.
2. In one triangle, $\angle J>\angle K$. Which side is longer: $\mathrm{KL}$ or $\mathrm{JL}$?

Show Answer

1) By the converse, $\angle A>\angle D$.

2) $\mathrm{KL}>\mathrm{JL}$ because it is opposite the larger angle $\angle J$.

Mini-summary 6 - use the Hinge Theorem forward to compare opposite sides from included angles. Use the converse to compare included angles from third sides. Use side–angle ordering inside one triangle to rank without calculation.

7) Multi-step reasoning with diagrams you can picture

Problem P1 - Adjustable stand comparison

Two adjustable stands each use arms of 60 cm and 40 cm meeting at a hinge. Stand A opens to 75°, Stand B opens to 50°. Which stand has the longer base between feet, and why?

Show Answer

Same side lengths with different included angles. The larger included angle at 75° produces the longer opposite base by the Hinge Theorem. So Stand A has the longer base.

Problem P2 - Same sides, different base lengths

Two models of a folding truss use bars of 28 cm and 19 cm. In Model X the base is 31 cm. In Model Y the base is 33 cm. Which model has the greater included angle at the hinge?

Show Answer

Two corresponding sides equal, longer base implies larger included angle by the converse. Model Y has the greater included angle.

Problem P3 - Inside-triangle ranking in a navigation corner

A route forms triangle $\Delta STU$ with $\angle S=38°$, $\angle T=67°$, $\angle U=75°$. Which segment is shortest, and which is longest?

Show Answer

Shortest side is opposite the smallest angle 38°, so $\mathrm{TU}$ is shortest. Longest is opposite 75°, so $\mathrm{ST}$ is longest.

Guiding question P4

Why do these comparisons not require computing any exact lengths with trigonometry or coordinates?

Show Answer

The theorems are comparative. They let you order sides and angles using only inequality logic and triangle structure, without needing precise measurements.

Mini-summary 7 - many design choices can be made with comparisons only. The hinge and side–angle rules avoid unnecessary calculations when you only need relative size.

8) Error analysis and common pitfalls

Pitfall 1 - Confusing included angle
In a triangle with sides AB and AC, the included angle is at A, not at B or C. Using a non-included angle with the Hinge Theorem gives wrong conclusions.

Pitfall 2 - Mixing within-triangle and between-triangle logic
Inside one triangle you use the side–angle order. Between two triangles with two corresponding sides equal, you use the Hinge Theorem or its converse.

Pitfall 3 - Assuming equality
The comparison theorems are strict. If angles are equal, the opposite sides are equal under the SAS-equality structure; if the third sides are equal with the same two sides, the included angles are equal.

Checkpoint E1 - Spot and fix

A student says: “Triangles with sides 12 and 9 will always have the longer third side when the non-included angle is larger.” What is wrong?

Show Answer

The Hinge Theorem compares the side opposite the included angle when two adjacent sides are fixed. A non-included angle does not control the opposite side in the SAS comparison.

Checkpoint E2 - Precision

Two triangles satisfy $\mathrm{AB}=\mathrm{DE}$, $\mathrm{AC}=\mathrm{DF}$, and $\mathrm{BC}=\mathrm{EF}$. What can you say about $\angle A$ and $\angle D$?

Show Answer

With two sides and the included angles compared via the converse, equality of the third sides implies the included angles are equal: $\angle A=\angle D$.

Mini-summary 8 - hinge requires the included angle. Equal third sides under matching two sides imply equal included angles.

9) Rich worked examples with step-by-step reasoning

Worked Example 7 - Compare opposite sides from included angles

Given $\Delta ABC$ and $\Delta DEF$ with $\mathrm{AB}=\mathrm{DE}=14$, $\mathrm{AC}=\mathrm{DF}=9$, and $\angle A=82°$, $\angle D=77°$. Determine which is larger: $\mathrm{BC}$ or $\mathrm{EF}$, and justify.

Show Answer

Same two sides correspond. Compare included angles: $\angle A>\angle D$. Hence by the Hinge Theorem, $\mathrm{BC}>\mathrm{EF}$.

Worked Example 8 - Use the converse to compare included angles

Now suppose the same side pairs are equal, but measurements show $\mathrm{EF}>\mathrm{BC}$. Which included angle is greater?

Show Answer

The triangle with the longer third side must have the larger included angle. Therefore $\angle D>\angle A$.

Worked Example 9 - Inside-triangle angle ranking and equality

In $\Delta GHI$, sides are $\mathrm{GH}=9$, $\mathrm{HI}=9$, $\mathrm{IG}=13$. Rank the angles and name the triangle type.

Show Answer

Equal sides $\mathrm{GH}$ and $\mathrm{HI}$ imply equal opposite angles $\angle I$ and $\angle G$. Longest side is $\mathrm{IG}=13$, so largest angle is $\angle H$. Order: $\angle H>\angle I=\angle G$. Triangle is isosceles.

Checkpoint W1 - Practice trio

1. Two frames have sides 22 cm and 15 cm. Base of Frame 1 is longer than base of Frame 2. Which included angle is larger?
2. In a triangle, $\angle P>\angle Q$. Compare $\mathrm{QR}$ and $\mathrm{PR}$.
3. Two triangles share sides 18 and 11. If one uses a 70° included angle and another uses a 40° included angle, which has the longer opposite side?

Show Answer

A) By the converse, the frame with the longer base has the larger included angle.

B) The side opposite the larger angle is longer, so $\mathrm{QR}>\mathrm{PR}$.

C) The 70° triangle has the longer opposite side by the Hinge Theorem.

Mini-summary 9 - worked examples demonstrate forward and converse use of the hinge comparison. Inside-triangle rankings follow instantly from side–angle pairing.

10) Guided investigations that you can run mentally

10.1 Varying the included angle gradually

Imagine two rigid bars of lengths 13 and 10. Start with a small included angle like 10°. Open to 30°, 60°, 100°, 150°. As the angle increases, the distance between the free tips increases as well. No numbers are needed to conclude comparison results, though formulas could compute exact distances.

Guiding question G1

If the angle increases from 30° to 80° while the bars remain 13 and 10, what happens to the opposite side, and which theorem guarantees your conclusion?

Show Answer

The opposite side gets longer. The Hinge Theorem guarantees that with two fixed sides, a larger included angle produces a longer opposite side.

Guiding question G2

If two designs both use sides 13 and 10 and you find that one has a longer opposite side, what must be true about their included angles?

Show Answer

The design with the longer opposite side has the larger included angle by the converse.

10.2 Internal ranking without all measurements

You can rank sides or angles from partial information. For instance, if you learn that $b>a$ in a triangle with side–angle pairing, then you instantly know $\angle B>\angle A$ without knowing angle values.

Guiding question G3

In $\Delta MNO$, suppose $\angle M<\angle N$. Compare $\mathrm{NO}$ and $\mathrm{MO}$.

Show Answer

The side opposite the larger angle is longer. Since $\angle N$ is larger than $\angle M$, the side opposite $\angle N$ is $\mathrm{MO}$, so $\mathrm{MO}>\mathrm{NO}$.

Mini-summary 10 - you can make strong comparison conclusions with partial data. Pair every side with its opposite angle to avoid confusion.

11) Applications with constraints and design choices

11.1 Picking a brace angle to fit a panel

An art panel is held by two arms of length 50 cm and 35 cm. The panel’s diagonal must span between the arm tips. You are free to choose the included angle by selecting among pin holes at 25°, 40°, 70°, and 110°.

Task A - Which choice maximizes the panel’s diagonal?

Show Answer

Same two side lengths for every choice. The largest included angle, 110°, yields the longest diagonal.

Task B - Which choice minimizes the diagonal?

Show Answer

The smallest included angle, 25°, yields the shortest diagonal. The hinge comparison is monotonic for fixed adjacent sides.

11.2 Designing a folding ladder stabilizer

Two equal-length stabilizer arms connect from a center hinge to floor pads. The base length between pads increases as the hinge opens.

Guiding question D1

If base slip must not exceed 1.6 m for safety, and the two arms are 1.2 m and 1.2 m, will opening the hinge beyond a certain angle violate the rule?

Show Answer

Yes. With equal arms, increasing the included angle increases the base. Past some angle, the base will exceed 1.6 m. The Hinge Theorem ensures the base grows with the angle when adjacent sides are fixed.

11.3 Game strategy with spacing triangles

A team drills keeping two distances fixed from a pivot. The passing lane length is the opposite side across from that pivot angle.

Task S - Coach’s instruction

Coach asks the team to widen the lane by 3 m without changing two fixed radii. What must change?

Show Answer

Increase the included angle at the pivot. With the same two sides, a larger included angle produces a longer lane.

Mini-summary 11 - practical choices often reduce to selecting an angle setting to get a longer or shorter opposite side. The comparisons give direction without computing exact numbers.

12) Mastery checkpoint set

MC1 - Two triangles have $\mathrm{AB}=\mathrm{DE}=26$ and $\mathrm{AC}=\mathrm{DF}=18$. If $\angle A<\angle D$, compare $\mathrm{BC}$ and $\mathrm{EF}$.

Show Answer

Smaller included angle yields shorter opposite side under matching adjacent sides. Thus $\mathrm{BC}<\mathrm{EF}$.

MC2 - With two triangles sharing two equal sides, the third sides are equal. What can you deduce about the included angles?

Show Answer

They are equal by the converse used in the equality case: if opposite sides match with the same two sides, the included angles must match.

MC3 - Inside a triangle, if $a>b$, what is the relationship between $\angle A$ and $\angle B$?

Show Answer

$\angle A>\angle B$.

MC4 - Two braces use equal bars 24 cm and 17 cm. Brace X has base 30 cm. Brace Y has base 28 cm. Which brace has the larger included angle?

Show Answer

The one with longer base has the larger included angle, so Brace X.

MC5 - In $\Delta CDE$, if $\angle C>\angle E$, compare $\mathrm{DE}$ and $\mathrm{CD}$.

Show Answer

Opposite the larger angle is the longer side. Opposite $\angle C$ is $\mathrm{DE}$, so $\mathrm{DE}>\mathrm{CD}$.

Mini-summary 12 - between triangles with two sides fixed, the included angle decides which opposite side is longer. Within a triangle, side and angle rankings mirror one another.

13) Mixed applied set with reasoning

A1 - Crane arms
A crane has arms of 7 m and 5 m. Setting A uses a boom angle of 40°, setting B uses 85°. Which hook-to-base distance is larger?

Show Answer

Setting B, since the included angle is larger with the same arms.

A2 - Folding chair base
Two chairs use legs of 0.9 m and 0.9 m. Chair 1 has base 1.2 m. Chair 2 has base 1.1 m. Which chair has a more open leg angle at the hinge?

Show Answer

Chair 1, because the longer base corresponds to the larger included angle when adjacent legs are equal.

A3 - City blocks
A walker travels 600 m east and 400 m north. Compare the direct corner-to-corner distance if the turn is increased from 45° to 90° between the same legs.

Show Answer

Larger turn angle between the same legs increases the direct distance, so the 90° version has the longer base.

A4 - Triangular kite frame
Two spars are 1.2 m and 1.6 m attached at the nose. The trailing edge length depends on the nose angle. Which nose setting increases the trailing edge: 30° or 75°?

Show Answer

75° yields a longer trailing edge.

Mini-summary 13 - different contexts tell the same story: larger included angle under fixed adjacent sides means a longer opposite span.

14) Synthesis tasks that combine Day 1 and Day 2

Day 1 taught you about feasibility and third-side bounds. Day 2 adds comparisons. Together, they let you check if a design is possible and then decide which configuration gives a larger or smaller span.

S1 - Given two arms 9 and 11, you know the third side satisfies $|11-9|<s<20$, so $2<s<20$. Within this feasible range, which angle choices make the span near the upper end?

Show Answer

Angles near straightening the arms produce a span approaching the sum, so larger included angles push $s$ toward the upper bound.

S2 - Two triangles with sides 13 and 8 must also respect third-side bounds from Day 1: $5<t<21$. If Triangle X has $t=20$ and Triangle Y has $t=8$, which included angle is larger?

Show Answer

Triangle X has the longer third side under the same two adjacent sides, so Triangle X has the larger included angle by the converse.

Mini-summary 14 - Day 1 determines what is allowed. Day 2 decides which is larger when options are allowed.

15) Deeper reasoning prompts

DR1 - Explain why the hinge comparison is reasonable using a thought experiment with a string stretched between two fixed nails as you rotate a board that sets the angle between the strings.

Show Answer

With two strings anchored at a common point and fixed in length, moving the included angle forces the free endpoints to move along arcs. As the angle increases, the distance between endpoints increases. That qualitative behavior is captured precisely by the Hinge Theorem.

DR2 - Why is it important that the angle you compare in SAS inequality is the included one, not an angle elsewhere in the triangle?

Show Answer

The side opposite the included angle is determined by how those two fixed sides open. If you compare a non-included angle, that angle does not control the side formed from the two fixed sides in the SAS setup.

DR3 - In a triangle where two angles are very close in size, what can you infer about the lengths of their opposite sides?

Show Answer

Their opposite sides must also be close in length. The side–angle order is continuous in this sense.

Mini-summary 15 - the included angle is the only angle that directly controls the opposite side when the adjacent sides are fixed. Near-equal angles imply near-equal opposite sides.

16) Consolidation with a structured practice block

Practice Set P - Try these. Use the right theorem or within-triangle pairing. Keep statements in MathML.

1. Two triangles have $\mathrm{AB}=\mathrm{DE}=18$, $\mathrm{AC}=\mathrm{DF}=12$, and $\angle D>\angle A$. Compare $\mathrm{BC}$ and $\mathrm{EF}$.
2. In $\Delta LMN$, $\angle L<\angle N$. Compare $\mathrm{MN}$ and $\mathrm{LM}$.
3. Two designs share arms 25 cm and 14 cm. The opposite spans are 34 cm and 30 cm. Which design has the larger included angle?
4. In $\Delta QRS$, sides are $\mathrm{QR}=6$, $\mathrm{RS}=9$, $\mathrm{SQ}=10$. Order the angles from largest to smallest.
5. Two triangles satisfy $\mathrm{AB}=\mathrm{DE}$, $\mathrm{AC}=\mathrm{DF}$, and $\mathrm{BC}=\mathrm{EF}$. Compare $\angle A$ and $\angle D$.

Show Answer

1) Larger included angle at D implies $\mathrm{EF}>\mathrm{BC}$.

2) Opposite larger angle is longer, so $\mathrm{LM}>\mathrm{MN}$.

3) The design with span 34 cm has the larger included angle by the converse.

4) Largest side is 10 opposite $\angle R$. Next is 9 opposite $\angle Q$. Smallest is 6 opposite $\angle S$. Order: $\angle R>\angle Q>\angle S$.

5) Equal third sides under equal pairs of sides force equal included angles: $\angle A=\angle D$.

Mini-summary 16 - you can now handle forward comparisons, converse logic, and within-triangle ranking. Always verify you are comparing the included angle when using the hinge ideas.

17) Reflection prompt inside Explore

Take one minute to write in your notebook: describe a real object with two fixed bars and a hinge. Explain how increasing the hinge angle changes a specific distance, naming which theorem supports your claim.

Show Answer

Example: A folding camera arm with fixed-length links. Increasing the elbow angle increases the distance from base to camera. This follows from the Hinge Theorem because the two adjacent sides are fixed and the included angle is larger, so the opposite side is longer.

18) References

• BYJU’S - SAS Inequality and hinge-type comparisons.
• Math is Fun - Triangle inequality family and angle–side relationships.
• MathBitsNotebook - Inequalities in one triangle and two triangles.
• Big Ideas Math - Inequalities in two triangles section.
• Encyclopedia Britannica - Triangle inequalities and comparison facts.

Worked Example 1 – Compare third sides (Hinge Theorem)

Two triangles satisfy $\mathrm{AB}=\mathrm{DE}=12$ and $\mathrm{AC}=\mathrm{DF}=7$. The included angles are $\angle A=64°$ and $\angle D=52°$. Decide which is longer, $\mathrm{BC}$ or $\mathrm{EF}$, and justify.

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Step 1 - Check SAS set-up: $\mathrm{AB}=\mathrm{DE}$, $\mathrm{AC}=\mathrm{DF}$. Angles compared are included.

Step 2 - Compare included angles: $\angle A>\angle D$.

Step 3 - Apply Hinge Theorem: larger included angle implies longer opposite side.

Conclusion - $\mathrm{BC}>\mathrm{EF}$.

Worked Example 2 – Compare included angles (Converse)

Two triangles have $\mathrm{AB}=\mathrm{DE}=9$, $\mathrm{AC}=\mathrm{DF}=14$, and third sides $\mathrm{BC}=19.2$, $\mathrm{EF}=18.7$. Which included angle is larger, $\angle A$ or $\angle D$?

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Longer third side is $\mathrm{BC}$. By the converse of the Hinge Theorem, $\angle A>\angle D$.

Worked Example 3 – Order sides from angles

In $\Delta RST$, $\angle R=48°$, $\angle S=64°$, $\angle T=68°$. Order $\mathrm{RS}$, $\mathrm{ST}$, $\mathrm{TR}$ from longest to shortest.

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Longest side opposite 68° is $\mathrm{RS}$, next opposite 64° is $\mathrm{RT}$, shortest opposite 48° is $\mathrm{ST}$. So $\mathrm{RS}>\mathrm{RT}>\mathrm{ST}$.

Worked Example 4 – Real-world hinge: frame span

A folding frame uses bars of $30$ cm and $18$. Setting X uses $45°$ as the included angle; Setting Y uses $95°$. Which setting gives a longer tip-to-tip span?

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Same adjacent sides, and $95°>45°$. By the Hinge Theorem, the larger included angle yields the longer opposite side. Setting Y produces a longer span.

Worked Example 5 – Equality case

Two triangles have $\mathrm{AB}=\mathrm{DE}=16$, $\mathrm{AC}=\mathrm{DF}=21$, and $\mathrm{BC}=\mathrm{EF}=25$. Compare the included angles $\angle A$ and $\angle D$.

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Equal third sides with equal adjacent sides imply equal included angles: $\angle A=\angle D$.

1. Two triangles satisfy $\mathrm{AB}=\mathrm{DE}=13$ and $\mathrm{AC}=\mathrm{DF}=9$. If $\angle A=72°$ and $\angle D=68°$, which is longer: $\mathrm{BC}$ or $\mathrm{EF}$?
   Show Answer

   $\angle A>\angle D$ with the same adjacent sides implies $\mathrm{BC}>\mathrm{EF}$ by the Hinge Theorem.
2. With $\mathrm{AB}=\mathrm{DE}=11$, $\mathrm{AC}=\mathrm{DF}=11$, suppose $\mathrm{BC}=18.2$ and $\mathrm{EF}=18.2$. Compare $\angle A$ and $\angle D$.
   Show Answer

   Equal third sides under equal adjacent sides force equal included angles: $\angle A=\angle D$.
3. In $\Delta ABC$, sides are $\mathrm{AB}=7$, $\mathrm{BC}=10$, $\mathrm{CA}=9$. Order the angles from largest to smallest.
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   Longest side is $\mathrm{BC}$ → largest angle $\angle A$. Next side $\mathrm{CA}$ → next angle $\angle B$. Shortest side $\mathrm{AB}$ → smallest angle $\angle C$. So $\angle A>\angle B>\angle C$.
4. In $\Delta PQR$, angles are $\angle P=50°$, $\angle Q=61°$, $\angle R=69°$. Order the sides from longest to shortest.
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   Largest angle $\angle R$ → longest side $\mathrm{PQ}$. Next angle $\angle Q$ → next side $\mathrm{PR}$. Smallest angle $\angle P$ → shortest side $\mathrm{QR}$. Thus $\mathrm{PQ}>\mathrm{PR}>\mathrm{QR}$.
5. Two triangles share $\mathrm{AB}=\mathrm{DE}=8$, $\mathrm{AC}=\mathrm{DF}=12$. If $\mathrm{BC}>\mathrm{EF}$, compare the included angles.
   Show Answer

   Longer third side implies larger included angle in that triangle: $\angle A>\angle D$.
6. Given $\mathrm{AB}=\mathrm{DE}=9$, $\mathrm{AC}=\mathrm{DF}=7$, a student claims: “Since $\angle B>\angle E$, then $\mathrm{BC}>\mathrm{EF}$.” Is this valid?
   Show Answer

   Not valid. Hinge compares sides opposite the included angle formed by the equal sides, so compare $\angle A$ and $\angle D$, not $\angle B$ and $\angle E$.
7. A linkage uses arms $15$ cm and $10$ cm. Configuration X has included angle $38°$; Configuration Y has $102°$. Which configuration creates the longer tip-to-tip span?
   Show Answer

   Same adjacent sides; larger included angle gives a longer opposite side. Configuration Y produces the longer span.
8. Two triangles share $\mathrm{AB}=\mathrm{DE}=18$, $\mathrm{AC}=\mathrm{DF}=12$, and $\mathrm{BC}=\mathrm{EF}$. What can you say about their included angles?
   Show Answer

   Equal third sides under equal adjacent sides imply equal included angles: $\angle A=\angle D$.
9. In $\Delta XYZ$, sides are $\mathrm{XY}=12$, $\mathrm{YZ}=5$, $\mathrm{ZX}=12$. Compare the angles $\angle Y$, $\angle Z$, and $\angle X$.
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   Equal sides imply $\angle Y=\angle Z$. The shortest side is $\mathrm{YZ}$ → smallest angle $\angle X$. Thus $\angle Y=\angle Z>\angle X$.
10. Two designs use bars $22$ cm and $17$ cm. Design A has opposite span $26.5$ cm; Design B has $31.0$ cm. Which design has the larger included angle?
    Show Answer

    The longer third side corresponds to the larger included angle. Since $31.0>26.5$, Design B has the larger included angle.

1. Two triangles have $\mathrm{AB}=\mathrm{DE}=12$ and $\mathrm{AC}=\mathrm{DF}=9$. If $\angle A=58°$ and $\angle D=73°$, compare $\mathrm{BC}$ and $\mathrm{EF}$.
   Show Answer

   $\angle D>\angle A$ gives $\mathrm{EF}>\mathrm{BC}$ by Hinge.
2. With $\mathrm{AB}=\mathrm{DE}=15$, $\mathrm{AC}=\mathrm{DF}=11$, suppose $\mathrm{BC}=19.4$ and $\mathrm{EF}=18.8$. Compare $\angle A$ and $\angle D$.
   Show Answer

   Longer third side → larger included angle: $\angle A>\angle D$.
3. In $\Delta PQR$, sides are $\mathrm{PQ}=14$, $\mathrm{QR}=10$, $\mathrm{RP}=12$. Order the angles from largest to smallest.
   Show Answer

   Longest side $\mathrm{PQ}$ → largest $\angle R$, next $\angle Q$, smallest $\angle P$.
4. In $\Delta ABC$, $\angle A=47°$, $\angle B=83°$, $\angle C=50°$. Order $a$, $b$, $c$ from longest to shortest.
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   $b>c>a$.
5. Decide if Hinge applies: equalities are $\mathrm{AB}=\mathrm{DE}$, $\mathrm{BC}=\mathrm{EF}$. Student compares $\angle B$ and $\angle E$ to conclude about $\mathrm{AC}$ vs $\mathrm{DF}$.
   Show Answer

   Not valid SAS-at-included-angle setup, so Hinge does not apply.
6. With $\mathrm{AB}=\mathrm{DE}=20$, $\mathrm{AC}=\mathrm{DF}=13$, and $\mathrm{BC}=\mathrm{EF}$, compare the included angles.
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   $\angle A=\angle D$.
7. In $\Delta JKL$, suppose $\angle J>\angle K$. Compare $\mathrm{KL}$ and $\mathrm{JL}$.
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   $\mathrm{KL}>\mathrm{JL}$.
8. A pivoted arm has links 28 cm and 16 cm. X uses $44°$; Y uses $92°$. Which has the longer span?
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   Y, by Hinge.
9. Two triangles share $\mathrm{AB}=\mathrm{DE}=17$ and $\mathrm{AC}=\mathrm{DF}=10$. If $\angle A<\angle D$, compare $\mathrm{BC}$ and $\mathrm{EF}$.
   Show Answer

   $\mathrm{BC}<\mathrm{EF}$.
10. In $\Delta STU$, sides are $\mathrm{ST}=9$, $\mathrm{TU}=14$, $\mathrm{US}=9$. Order $\angle S$, $\angle T$, $\angle U$.
    Show Answer

    $\angle S>\angle U=\angle T$.
11. Enough information? You know $\mathrm{AB}=\mathrm{DE}$, unknown $\mathrm{AC}$ and $\mathrm{DF}$. You measure $\mathrm{BC}>\mathrm{EF}$. Can you conclude $\angle A>\angle D$?
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    No. Converse needs two corresponding side pairs equal.
12. Inside triangle: if $\mathrm{MN}>\mathrm{NP}$, compare $\angle P$ and $\angle M$.
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    $\angle P>\angle M$.
13. Bars are 26 cm and 19 cm. Option A uses $39°$. Option B uses $101°$. Which yields a longer opposite side?
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    Option B.
14. Two frames use bars 12.5 cm and 9.0 cm. Bases: 18.7 cm and 17.9 cm. Which has the larger included angle?
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    The one with 18.7 cm base.
15. (a) In $\Delta XYZ$, angles are $\angle X=42°$, $\angle Y=42°$, $\angle Z=96°$. Order the sides. (b) Two other triangles share equal sides 18 cm and 12 cm; the one with longer third side $t$ has which included angle comparison?
    Show Answer

    (a) $\mathrm{XY}>\mathrm{YZ}=\mathrm{XZ}$. (b) That triangle has the larger included angle by the converse.

1) Monotonic “hinge” proof with a function of the included angle

You have two fixed side lengths $a$ and $b$. Let the included angle be $\theta$ and the opposite side be $c$:

$$c^2=a^2+b^2-2ab\cos \left(\theta \right)$$

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Since $\cos \left(\theta \right)$ is decreasing on $(0,180°)$, the term $-2ab\cos \left(\theta \right)$ increases, so $c^2$ increases and thus $c$ increases.

2) Design window: convert a required span range into an angle range

Two bars have lengths $a=12$ cm and $b=20$ cm. The opposite span must lie in $[18,26]$ cm. Find the allowable interval for $\theta$.

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$$\cos \left(\theta \right)=\frac{a^2+b^2-d^2}{2ab}$$ → angle range $\theta \in [62.7°,106.0°]$.

3) Quantify how much larger the angle is when the third side grows

Same adjacent bars $a=9$ cm and $b=14$. Spans 15 cm and 17 cm. Estimate the angle increase.

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Using cosine relation gives approximately $\Delta \theta \approx 14.6°$.

4) Synthetic proof idea (no calculator)

Give a construction-based argument for the hinge comparison without trigonometry.

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Construct a triangle with same adjacent sides and rotate a ray to match the larger included angle; the opposite side extends beyond, forcing it to be longer, matching the SAS-inequality idea.

5) Parametric table: how span grows with the angle

Bars $a=9$, $b=12$. Use $d^2=225-216\cos \left(\theta \right)$ to fill values for $\theta =30°$, $45°$, $60°$, $75°$, $90°$, $120°$.

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Approximate results: 6.16, 8.50, 10.82, 13.00, 15.00, 18.25 (cm), increasing with $\theta$.

3–2–1 Reflection

1. Three ideas I can now explain clearly
2. Two questions I still have
3. One real-life situation where I will apply today’s ideas
   • Describe the objects (two fixed sides, included angle).
   • Explain what changes and what becomes longer or shorter.
   • State your conclusion using an inequality in MathML if possible.