Question

directions: order the sides from least to greatest for 17 - 18, then greatest to least for 19 - 20.
17.
18.
19.
20.

Explanation:

Step1: Find the third angle for problem 17

In triangle \(ABC\), using the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)). Given \(\angle A = 92^{\circ}\), \(\angle B=51^{\circ}\), then \(\angle C=180-(92 + 51)=37^{\circ}\).
In a triangle, the side opposite the smaller angle is shorter.
The angles are \(\angle C = 37^{\circ}\), \(\angle B = 51^{\circ}\), \(\angle A=92^{\circ}\).
The sides opposite these angles are \(AB\), \(AC\), \(BC\) respectively.
So the order from least to greatest: \(AB\lt AC\lt BC\)

Step2: Find the third angle for problem 18

In triangle \(MLN\), using the angle - sum property of a triangle (\(\angle M+\angle N+\angle L = 180^{\circ}\)). Given \(\angle M = 78^{\circ}\), \(\angle N = 48^{\circ}\), then \(\angle L=180-(78 + 48)=54^{\circ}\).
The angles are \(\angle N = 48^{\circ}\), \(\angle L = 54^{\circ}\), \(\angle M=78^{\circ}\).
The sides opposite these angles are \(ML\), \(MN\), \(LN\) respectively.
So the order from least to greatest: \(ML\lt LN\lt MN\)

Step3: Find the third angle for problem 19

In triangle \(RST\), using the angle - sum property of a triangle (\(\angle R+\angle S+\angle T = 180^{\circ}\)). Given \(\angle S = 60^{\circ}\), \(\angle T = 57^{\circ}\), then \(\angle R=180-(60 + 57)=63^{\circ}\).
The angles are \(\angle T = 57^{\circ}\), \(\angle S = 60^{\circ}\), \(\angle R=63^{\circ}\).
The sides opposite these angles are \(RS\), \(RT\), \(ST\) respectively.
So the order from greatest to least: \(ST\gt RT\gt RS\)

Step4: Find the third angle for problem 20

The exterior angle at the non - labeled vertex adjacent to \(28^{\circ}\) is \(180 - 28=152^{\circ}\). In triangle \(HJK\), using the angle - sum property of a triangle. Let's first find \(\angle H\): \(\angle H=180-(118 + 28)=34^{\circ}\)
The angles are \(\angle H = 34^{\circ}\), \(\angle J = 28^{\circ}\), \(\angle K = 118^{\circ}\).
The sides opposite these angles are \(JK\), \(HK\), \(HJ\) respectively.
So the order from greatest to least: \(HJ\gt HK\gt JK\)

Answer:

1. \(AB\lt AC\lt BC\)
2. \(ML\lt LN\lt MN\)
3. \(ST\gt RT\gt RS\)
4. \(HJ\gt HK\gt JK\)