In this explainer, we will learn how to form inequalities involving the lengths of the sides of a triangle given the measures of the angles of the triangle.
In an isosceles triangle, we can recall that the angles opposite the sides of equal lengths are of equal measures. The converse to this statement is also true: if two angles in a triangle have equal measures, then the sides opposite these angles have the same lengths. We can use this idea to consider what happens if we have two angles in a triangle with different measures.
We can start by considering an isosceles triangle. We know that the angles opposite the sides of equal lengths have equal measures. We can compare the lengths and measures of the angles by altering the base of the triangle. If we decrease one of the angles at the base (while leaving the third angle unchanged), then the other angle at the base must increase since the sum of the measures of the interior angles is constant at .
We can also see that doing this decreases the length of the side opposite the smaller angle. In other words, the smaller angle is opposite the smaller side. We can follow the same process if we increase the measures of one of the angles at the base.
Since we have increased the measure of one angle, the other angle must decrease in measure. We can also see that we have increased the length of the side opposite the angle of the larger measure.
This result holds true in general and is called the side comparison theorem in triangles.
Theorem: Side Comparison Theorem in Triangles
If we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle. In particular, consider .
If we know that , then .
It is also worth noting that we can apply this result with the inequality reversed. In particular, if we have a triangle such that two sides have unequal lengths, then the side opposite the smaller angle is shorter than the side opposite the angle with the larger measure.
Letβs see an example of applying this result to construct an inequality of the side lengths of a triangle using its angle measures.
Example 1: The Relation between Sides and Corresponding Angles in a Triangle
Arrange the side lengths of from least to greatest.
Answer
We are given two of the measures of the interior angles in a triangle and asked to use this to arrange the side lengths in ascending order. We can do this by recalling that the side comparison theorem for triangles tells us that, if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
We know that , so we can conclude that the side opposite is longer than the side opposite . Thus,
We can find the measure of by recalling that the sum of the measures of the interior angles in a triangle is . Therefore,
We can substitute the measures of and to get
We then solve for :
We can now see that , so the side opposite must be longer than the side opposite . Thus,
We can combine these two inequalities into a compound inequality:
Finally, we are told to write the side lengths from least to greatest. This means we start with the shortest side and end with the longest side.
We can reverse the compound inequality to get the sides listed from least to greatest in length:
In our next example, we will use the side comparison theorem in triangles to determine which of two sides in a triangle is longer by using the measures of two exterior angles in the triangle.
Example 2: Completing Inequalities for the Side Lengths of a Triangle
From the figure, how do and compare?
Answer
We want to compare the lengths of two sides of triangle . We can do this by recalling that the side comparison theorem in triangles tells us that, if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
This means we can compare the lengths of the sides of the triangle by comparing the measures of the angles opposite the sides. We can mark the angles opposite each side in the diagram.
We can see that and combine to make a straight angle, so their measures sum to give . Thus,
We can see in the diagram that , so we can substitute this value into the equation to get
We can then solve for
We can now directly find by noting that is the opposite exterior angle in to and .
We can recall that the measure of an exterior angle in a triangle is equal to the sum of the measures of the opposite interior angles. Thus,
We know that and , so we can substitute these values into the equation to get
We can subtract from both sides of the equation to get
We can add this onto the diagram.
We see that the angle opposite has a larger measure than the angle opposite ; therefore, it must be the longer side.
Hence, .
In our next example, we will compare four pairs of side lengths in triangles by using the side comparison theorem in triangles.
Example 3: Completing Inequalities for the Side Lengths of a Triangle
Consider the given diagram.
Fill in the blanks in the following statements using , or .
Answer
We first recall the side comparison theorem in triangles that tells us that if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
This means we can compare the lengths of the sides of the triangle by comparing the measures of the angles opposite the sides.
Part 1
We want to compare the lengths of and . We can start by highlighting these line segments and the angles opposite them on the diagram.
We can note that , so the side opposite the right angle must be longer than the side opposite the angle of measure .
Hence,
Part 2
We can follow the same process to compare the lengths of and . We highlight both line segments on the diagram as well as the angles opposite each side.
We can determine the measure of the angle opposite by adding the measures of the two angles that combine to form . We have
Since both angles have the same measures, we can conclude that is isosceles with .
Part 3
To compare the lengths of and , we can first highlight both line segments on the diagram as well as the angles opposite each side.
Since the measure of the angle opposite is smaller than the measure of the angle opposite , we can conclude that is shorter than .
Hence, .
Part 4
Finally, to compare the lengths of and , we highlight both line segments on the diagram as well as the angles opposite each side.
It appears as though the measure of is greater than that of . However, for due diligence, we should verify this. We can find the measure of this angle by using the fact that the sum of the measures of the interior angles in a triangle is . Applying this to , we have
Substituting the known angle measures into the equation gives
We can then solve for :
We can now see that , so the side opposite is longer than the side opposite .
Hence, .
In the previous example, we saw that the side opposite the right angle in a right triangle was longer than the other side. We can show that this result is true in general.
We know that the sum of the measures of the interior angles in a triangle is . So, in this triangle, we have
Substituting into the equation and rearranging gives us
Since the angle measures must be positive, we can note that both of these angles have measures less than . In particular, this tells us that has the largest measure in the triangle. Therefore, the side opposite the right angle in a right triangle is the longest side. We call this the hypotenuse of the right triangle.
This agrees with our knowledge of the Pythagorean theorem. We recall that this tells us that, in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two other sides. In our right triangle above, this is
Since , and are all lengths, they are all positive, so we can note that and . Thus, the hypotenuse must be the longest side.
In our next example, we will prove an inequality involving the lengths of lines in a geometric construct.
Example 4: Proving an Inequality Involving the Side Lengths of a Triangle
From the figure, fill in the blank with either , or : .
Answer
We want to compare the lengths of two sides of a triangle. We can do this by recalling that the longer side will be opposite the angle with the larger measure. Letβs start by highlighting the sides in the given inequality and the angles opposite each side. We will call the angle measures and as shown.
We can see that the angles at make a straight angle, so their measures sum to give . Thus,
Solving for gives
We can find by noting that it is an alternate interior angle to another angle in a transversal of parallel lines.
We can then find by using the fact that the sum of the measures in a straight angle is or by noting that is the corresponding angle of the angle of measure in the diagram.
Since the angle opposite has a larger measure than the angle opposite , we can conclude that is longer than . Hence,
We can prove the side comparison theorem by considering a triangle in which .
We can first show that and do not have the same length; if they did, then would be an isosceles triangle since it has two sides of equal lengths. We can recall that the angles opposite the sides of equal lengths must have equal measures. This would mean that , but we are told that , so this cannot be true. Hence, and must have unequal lengths.
We can also show that cannot be longer than . We recall that the angle comparison theorem in triangles tells us that if we have a triangle such that two sides have unequal lengths, then the angle opposite the longer side has a larger measure than the angle opposite the shorter side.
If we assume that is longer than , then this theorem tells us that the angle opposite must have a larger measure than the angle opposite . In other words, we would have . However, we are told that , so this cannot be the case. Hence, cannot be longer than .
Since and cannot have the same length and cannot be longer than , there is only one final possibility; is longer than .
In our next example, we will use the side comparison theorem to compare lengths in a given construct.
Example 5: Comparing Side lengths in a Triangle Using the Side Comparison Theorem
Consider the figure shown.
Fill in the blank with , or : .
Answer
From the diagram, we can note that , so triangle is an isosceles triangle. The angles opposite the sides of equal lengths must be congruent. We can add these congruent angles onto the diagram.
We can now recall that the side comparison theorem in triangles tells us that if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
We see that is opposite and is opposite . We can also see from the diagram that has a larger measure than since it is the combination of two angles, one of which has an equal measure to :
We can conclude that the side opposite the angle with a lower measure is shorter. Hence,
We can use the angle comparison theorem in triangles to prove many useful geometric results. For example, consider and line segment such that .
We can show that is the shortest line segment from to . Consider any other point .
We see that is a right triangle, so its hypotenuse is the longest side. Thus,
This is true for any point on , so is the shortest line segment from to .
This holds for any perpendicular line from a point to a straight line.
Property: Perpendicular from a Point to a Line
The perpendicular line segment connecting a line and a point is the shortest line segment connecting that line and that point.
Letβs finish by recapping some of the important points from this explainer.
Key Points
- If we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
- In a right triangle, the hypotenuse is the longest side since it is opposite the right angle, which is the angle with the largest measure.
- The perpendicular line segment drawn from one side of a triangle to the opposite vertex is shorter than any other line segment drawn from the same side to the same vertex.