Question

1. list the sides in order from shortest to longest.
2. list two combinations of segment lengths that will not produce triangles.
3. two airplanes leave the same airport heading in different directions. after 2 hours, one airplane has traveled 710 miles and the other has traveled 640 miles. describe the range of distances that represent how far apart the two airplanes can be at this time.

Explanation:

Step1: Recall angle - side relationship in a triangle

In a triangle, the side opposite the smallest angle is the shortest and the side opposite the largest angle is the longest.

Step2: Identify the smallest and largest angles

In $\triangle XYZ$, $\angle Z = 37^{\circ}$, $\angle X=45^{\circ}$, $\angle Y = 98^{\circ}$. The smallest angle is $\angle Z$, the middle - sized angle is $\angle X$, and the largest angle is $\angle Y$.

Step3: List the sides in order

The side opposite $\angle Z$ is $\overline{XY}$, the side opposite $\angle X$ is $\overline{YZ}$, and the side opposite $\angle Y$ is $\overline{XZ}$. So the sides in order from shortest to longest are $\overline{XY}$, $\overline{YZ}$, $\overline{XZ}$.

Step4: Recall the triangle - inequality theorem for non - triangle combinations

The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. For example, if we have side lengths $a$, $b$, and $c$, then $a + b>c$, $a + c>b$, and $b + c>a$. Combinations like $1, 2, 4$ (since $1+2<4$) and $2, 3, 5$ (since $2 + 3=5$) will not form triangles.

Step5: Use the triangle - inequality for the airplane problem

Let the distances traveled by the two airplanes be $a = 710$ miles and $b = 640$ miles. The distance $d$ between them forms a triangle with sides $a$ and $b$. By the triangle - inequality, $|a - b|\leq d\leq a + b$. So $|710-640|\leq d\leq710 + 640$, which simplifies to $70\leq d\leq1350$.

Answer:

1. $\overline{XY}$, $\overline{YZ}$, $\overline{XZ}$
2. 1, 2, 4; 2, 3, 5
3. The two airplanes are between 70 miles and 1350 miles apart.