Question

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

Explanation:

Problem 17: Triangle \( \triangle ABC \)

Step 1: Find the third angle

In a triangle, the sum of angles is \( 180^\circ \). Given \( \angle A = 92^\circ \), \( \angle B = 51^\circ \), so \( \angle C = 180 - 92 - 51 = 37^\circ \).

Step 2: Order angles from least to greatest

\( \angle C (37^\circ) < \angle B (51^\circ) < \angle A (92^\circ) \)

Step 3: Relate angles to opposite sides

In a triangle, the larger the angle, the longer the opposite side. Opposite sides: \( \angle A \) opposite \( BC \), \( \angle B \) opposite \( AC \), \( \angle C \) opposite \( AB \). So sides from least to greatest: \( AB < AC < BC \)

Problem 18: Triangle \( \triangle LMN \)

Step 1: Find the third angle

Sum of angles is \( 180^\circ \). \( \angle M = 72^\circ \), \( \angle N = 48^\circ \), so \( \angle L = 180 - 72 - 48 = 60^\circ \).

Step 2: Order angles from least to greatest

\( \angle N (48^\circ) < \angle L (60^\circ) < \angle M (72^\circ) \)

Step 3: Relate angles to opposite sides

Opposite sides: \( \angle M \) opposite \( LN \), \( \angle N \) opposite \( LM \), \( \angle L \) opposite \( MN \). So sides from least to greatest: \( LM < MN < LN \)

Problem 19: Triangle \( \triangle RST \)

Step 1: Find the third angle

Sum of angles is \( 180^\circ \). \( \angle S = 60^\circ \), \( \angle T = 57^\circ \), so \( \angle R = 180 - 60 - 57 = 63^\circ \).

Step 2: Order angles from greatest to least

\( \angle R (63^\circ) > \angle S (60^\circ) > \angle T (57^\circ) \)

Step 3: Relate angles to opposite sides

Opposite sides: \( \angle R \) opposite \( ST \), \( \angle S \) opposite \( RT \), \( \angle T \) opposite \( RS \). So sides from greatest to least: \( ST > RT > RS \)

Problem 20: Triangle \( \triangle HJK \)

Answer:

s:

1. \( AB < AC < BC \)
2. \( LM < MN < LN \)
3. \( ST > RT > RS \)
4. \( HJ > JK > HK \)