Question

section 6-2 practice
independent practice

1. list the angles and sides of each triangle in order from smallest to largest.

(there are two triangle figures, one with vertices s, t, and a top vertex, angles at s is 60°, at t is 54°; the other with vertices p, q, r, sides pr = 12, rq = 18, pq = 20)

Explanation:

Step1: Find ∠R in △RST

First, calculate the missing angle using the triangle angle sum (180°).
$\angle R = 180^\circ - 60^\circ - 54^\circ = 66^\circ$

Step2: Order angles (△RST, smallest to largest)

Compare the three angles: $54^\circ < 60^\circ < 66^\circ$, so map to vertices: $\angle T < \angle S < \angle R$

Step3: Order sides (△RST, smallest to largest)

Sides correspond to opposite angles: side opposite ∠T is $\overline{RS}$, opposite ∠S is $\overline{RT}$, opposite ∠R is $\overline{ST}$. So $\overline{RS} < \overline{RT} < \overline{ST}$

Step4: Order sides (△PQR, smallest to largest)

Compare the given side lengths: $12 < 18 < 20$, so $\overline{PR} < \overline{RQ} < \overline{PQ}$

Step5: Order angles (△PQR, smallest to largest)

Angles correspond to opposite sides: angle opposite $\overline{PR}$ is $\angle Q$, opposite $\overline{RQ}$ is $\angle P$, opposite $\overline{PQ}$ is $\angle R$. So $\angle Q < \angle P < \angle R$

Answer:

For △RST:

Angles (smallest to largest): $\angle T$, $\angle S$, $\angle R$
Sides (smallest to largest): $\overline{RS}$, $\overline{RT}$, $\overline{ST}$

For △PQR:

Sides (smallest to largest): $\overline{PR}$, $\overline{RQ}$, $\overline{PQ}$
Angles (smallest to largest): $\angle Q$, $\angle P$, $\angle R$