# Chapter 5 - Relationships Within Triangles - 5-6 Inequalities in One Triangle - Practice and Problem-Solving Exercises - Page 329: 17

$\overline{TU} < \overline{UV} < \overline{TV}$

#### Work Step by Step

First, let's figure out the measures of all the angles in the triangle given. We already have the measures of two of the angles. One angle is $30^{\circ}$ and the other is $90^{\circ}$ because it is a right angle. We use the triangle-sum theorem to figure out the measure of the third angle: $m \angle T = 180 - (90 + 30)$ Evaluate what is in parentheses first, according to order of operations: $m \angle H = 180 - (120)$ Subtract to solve: $m \angle H = 60$ Let's now put the measures of the angles in order from smallest to largest: $30^{\circ} < 60^{\circ} < 90^{\circ}$ Now, put the angles corresponding to these measures in order from smallest to largest: $\angle V < \angle T < \angle U$ From Theorem 5-11, which states that the longer side of a triangle is opposite the largest angle, we can write out the order of sides from shortest to longest: $\overline{TU} < \overline{UV} < \overline{TV}$

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