Question

read 7.3 inequalities in one triangle (textbook pgs. 426 - 429)
angle - side relationships when the sides of triangles are not congruent, there is a relationship between the sides and angles of the triangles.

• if one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.
• if one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.

example 1 list the angles in order from smallest to largest measure.
(figure of triangle rst with sides rs = 6 cm, st = 9 cm, rt = 7 cm)
∠t, ∠r, ∠s
example 2 list the sides in order from shortest to longest.
(figure of triangle abc with angles ∠a = 20°, ∠b = 125°, ∠c = 35°)
\\(\overline{cb}\\), \\(\overline{ab}\\), \\(\overline{ac}\\)
exercises
list the angles and sides in order from smallest to largest.

1. (figure of triangle rts with sides rt = 48 cm, ts = 36 cm, sr = 21.7 cm)
2. (figure of triangle rst with angles ∠s = 30°, ∠r = 50°, ∠t = 40°)
3. (figure of triangle abc with sides ab = 3.8, bc = 4.3, ac = 4.0)
4. (figure of triangle sut with sides su = 14, ut = 5, st = 11)
5. (figure of triangle abc with sides ab = 5, ac = 4, bc = 8)
6. (figure of triangle pqr with sides pr = 12, pq = 20, rq =?)

Explanation:

Step1: For Exercise 1: Match sides to opposite angles

Sides: $TS=35\ \text{cm}$, $SR=21.7\ \text{cm}$, $TR=41\ \text{cm}$
Opposite angles: $\angle R$ (opposite $TS$), $\angle T$ (opposite $SR$), $\angle S$ (opposite $TR$)
Order of sides (shortest to longest): $SR < TS < TR$
Order of angles (smallest to largest): $\angle T < \angle R < \angle S$

Step2: For Exercise 2: Match angles to opposite sides

Angles: $m\angle S=90^\circ$, $m\angle R=60^\circ$, $m\angle T=40^\circ$
Opposite sides: $\overline{RT}$ (opposite $\angle S$), $\overline{ST}$ (opposite $\angle R$), $\overline{SR}$ (opposite $\angle T$)
Order of angles (smallest to largest): $\angle T < \angle R < \angle S$
Order of sides (shortest to longest): $\overline{SR} < \overline{ST} < \overline{RT}$

Step3: For Exercise 3: Match sides to opposite angles

Sides: $AB=3.8$, $BC=4.3$, $AC=4.0$
Opposite angles: $\angle C$ (opposite $AB$), $\angle A$ (opposite $BC$), $\angle B$ (opposite $AC$)
Order of sides (shortest to longest): $AB < AC < BC$
Order of angles (smallest to largest): $\angle C < \angle B < \angle A$

Step4: For Exercise 4: Match sides to opposite angles

Sides: $ST=11$, $TU=15$, $SU=14$
Opposite angles: $\angle U$ (opposite $ST$), $\angle S$ (opposite $TU$), $\angle T$ (opposite $SU$)
Order of sides (shortest to longest): $ST < SU < TU$
Order of angles (smallest to largest): $\angle U < \angle T < \angle S$

Step5: For Exercise 5: Match sides to opposite angles

Sides: $AC=4$, $AB=5$, $BC=8$
Opposite angles: $\angle B$ (opposite $AC$), $\angle C$ (opposite $AB$), $\angle A$ (opposite $BC$)
Order of sides (shortest to longest): $AC < AB < BC$
Order of angles (smallest to largest): $\angle B < \angle C < \angle A$

Step6: For Exercise 6: Match sides to opposite angles

Sides: $PR=12$, $PQ=20$, $RQ=\text{unknown, but use side order}$
Opposite angles: $\angle Q$ (opposite $PR$), $\angle R$ (opposite $PQ$), $\angle P$ (opposite $RQ$)
First, find $RQ$ via triangle inequality (not needed for order): $20-12 < RQ < 20+12$, so $8 < RQ < 32$. Order of given sides: $PR < PQ$, so $RQ$ is between them (since $PQ$ is longest)
Order of sides (shortest to longest): $PR < RQ < PQ$
Order of angles (smallest to largest): $\angle Q < \angle P < \angle R$

Answer:

1. Angles: $\boldsymbol{\angle T < \angle R < \angle S}$; Sides: $\boldsymbol{\overline{SR} < \overline{TS} < \overline{TR}}$
2. Angles: $\boldsymbol{\angle T < \angle R < \angle S}$; Sides: $\boldsymbol{\overline{SR} < \overline{ST} < \overline{RT}}$
3. Angles: $\boldsymbol{\angle C < \angle B < \angle A}$; Sides: $\boldsymbol{\overline{AB} < \overline{AC} < \overline{BC}}$
4. Angles: $\boldsymbol{\angle U < \angle T < \angle S}$; Sides: $\boldsymbol{\overline{ST} < \overline{SU} < \overline{TU}}$
5. Angles: $\boldsymbol{\angle B < \angle C < \angle A}$; Sides: $\boldsymbol{\overline{AC} < \overline{AB} < \overline{BC}}$
6. Angles: $\boldsymbol{\angle Q < \angle P < \angle R}$; Sides: $\boldsymbol{\overline{PR} < \overline{RQ} < \overline{PQ}}$