![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-6.png\")
<\/figure>\n\n\n\nThis theorem applies to all types of triangles, so this shows how important this theorem is as it opens a wide range of properties and applications. Take a look at the two triangles shown below \u2013 [katex]\\Delta ABC[\/katex] is a scalene triangle and [katex]\\Delta MNP[\/katex] is a right triangle.<\/p>\n\n\n\n As we have learned, the triangle sum theorem states that the sum of the interior angles of each of these triangles will be 180o<\/sup>. Test this out for yourself by adding these angles and see the beauty of this theorem!<\/p>\n\n\n\n Knowing this interesting property about triangles opens a wide range of applications. Given two interior angles of a triangle, it is now possible to find the measure of the third interior angle. First, try out this problem by remembering a triangle\u2019s interior angles can only have one sum: 180o<\/sup>.<\/p>\n\n\n\n Which of the following triangles do not have valid interior angle measures?<\/p>\n\n\n\n As established earlier, the sum of any triangles\u2019 interior angles must be 180o<\/sup>. This means that if the interior angles do not add up to 180o<\/sup>, the triangle doesn\u2019t have valid interior angle measures. Use this fact when identifying the triangle we\u2019re looking for and begin by adding the interior angles for each triangle.<\/p>\n\n\n\n From the table, we can see that the interior angles of the first three triangles all each have a sum of 180o<\/sup>. However, in \u2206MLN, the sum of its interior angles is equal to 190o<\/sup>. This can\u2019t be possible since the triangle sum theorem covers all triangles, so the interior angles of this triangle aren\u2019t valid.<\/p>\n\n\n\n Before learning how to apply the triangle sum theorem and try out other problems, it\u2019s important to understand how it\u2019s possible that the interior angles of any triangle will always add up to 180o<\/sup>. Why don\u2019t we take a look at its proof first?<\/p>\n\n\n\n Proving the triangle sum theorem begins by constructing a line that passes through one of the vertices. This line must also be parallel to the side of the triangle facing opposite the vertex that the line passes through. Use the properties of parallel lines to show that the sum of the three interior angles is 180o<\/sup>. <\/p>\n\n\n\n Now, observe the angles of this figure starting with the ones near the vertex, A. The two angles, [katex]\\angle NAC[\/katex] and [katex]\\angle CAB[\/katex], form the larger angle, [katex]\\angle NAB[\/katex]. Since MN <\/em>is a line, the two angles that form it are linear pair of angles. Here are the relationships that have been established so far:<\/p>\n\n\n\n [katex]\\begin{aligned}\\angle NAB &= \\angle NAC + \\angle CAB\\,\\, (1)\\\\\\angle NAB + \\angle MAB &= 180\\degree\\,\\,(2)\\end{aligned}[\/katex]<\/p>\n\n\n\n Use the right-hand side of the first equation to rewrite the expression for [katex]\\angle NAB[\/katex] in the second equation.<\/p>\n\n\n\n [katex] \\begin{aligned}(\\angle NAC + \\angle CAB) + \\angle MAB &= 180\\degree \\end{aligned} [\/katex]<\/p>\n\n\n\n Recall that alternate interior angles of parallel lines cut by a transversal line will have equal measures. Now, since MN<\/em> and BC<\/em> are parallel lines, we have the following pairs of interior angles: [katex]\\angle MAB =\\angle ABC[\/katex] and [katex]\\angle NAC =\\angle ACB[\/katex]. Use these relationships to rewrite the previous equation so that the right-hand side will have all the interior angles of \u2206ABC.<\/p>\n\n\n\n [katex] \\begin{aligned}\\angle NAC + \\angle CAB + \\angle MAB &= 180\\degree\\\\ \\angle ACB +\\angle CAB +\\angle ABC &= 180\\degree\\end{aligned}[\/katex]<\/p>\n\n\n\n Hence, it has been shown that the interior angles of a triangle will always have a sum of 180o<\/sup>. Now that the theorem has been proven for all triangles, it\u2019s time to learn how to use this theorem to solve problems involving triangles and interior angles. <\/p>\n\n\n\n When using the triangle sum theorem, identify the given interior angles and subtract the sum of these angles from 180o<\/sup>. Since triangles\u2019 interior angles will always add up to 180o<\/sup>, the process of solving problems involving them will often require the triangle sum theorem. <\/p>\n\n\n\n Here are some steps to remember when solving problems using the triangle sum theorem:<\/p>\n\n\n\n Step 1:<\/strong> Identify the triangle\u2019s three interior angles.<\/p>\n\n\n\n Step 2: <\/strong>Equate their sum\u2019s expression to 180o<\/sup>.<\/p>\n\n\n\n Step 3: <\/strong>Solve the resulting equation to find the measure of the unknown interior angle. <\/p>\n\n\n\n Of course, the best way to master the triangle sum theorem is through practice. Work on the problems below to understand and appreciate this special property of interior angles within a triangle. <\/p>\n\n\n\n The two interior angles of a triangle have the following measures: 40o<\/sup> and 80o<\/sup>. What is the measure of the third interior angle?<\/p>\n\n\n\n The triangle sum theorem states that the sum of the three interior angles of the triangle is 180o<\/sup>, so the sum of 400<\/sup>, 80o<\/sup>, and the third angle (let it be x<\/em>) must also be 180o<\/sup>.<\/p>\n\n\n\n Write the sum of the triangles in terms of x<\/em> then equate the expression to 180o<\/sup>. This leads to an equation, so simplify it then solve for x<\/em>.<\/p>\n\n\n\n [katex] \\begin{aligned}40\\degree + 80\\degree + x\\degree &= 180\\degree\\\\ (120 + x)\\degree &= 180\\degree\\\\x &= 180 – 120\\\\x&= 60\\end{aligned} [\/katex]<\/p>\n\n\n\n This means that the third interior angle\u2019s measure is 60o<\/sup>. Another way of approaching this problem is by working backward. We simply subtract the first angle from 180o<\/sup> then take out the second interior angle from the difference. This leaves us with the same measure for the third interior angle- 60o<\/sup>.<\/p>\n\n\n\n This approach is helpful when working quickly on numerical angle measures. If algebraic expressions are involved, stick with the first method for a more systematic approach to solving the problem.<\/p>\n\n\n\n Given the right triangle, \u2206MNP, shown by the diagram below, what are the measures of [katex]\\angle NMP[\/katex] and [katex]\\angle MPN[\/katex]?<\/p>\n\n\n\n Right triangles will always have 90o<\/sup> as one of their interior angles. In fact, for \u2206MNP, [katex]\\angle MNP = 90\\degree[\/katex]. This means that the sum of the two remaining interior angles is 90o<\/sup>. Add the two angles then equate it with 90o<\/sup>.<\/p>\n\n\n\n [katex] \\begin{aligned}\\angle MNP + \\angle NMP + \\angle MPN &= 180\\degree\\\\90\\degree + x\\degree + (2x + 15)\\degree &= 180\\degree\\\\ x\\degree + (2x + 15)\\degree &= 90\\degree\\\\3x + 15&= 90\\\\3x &= 75\\\\x&= 25 \\end{aligned} [\/katex]<\/p>\n\n\n\n This means that [katex]\\angle NMP = x \\degree = 25\\degree[\/katex]. To find the measure of [katex]\\angle PMN[\/katex], substitute [katex]x =25[\/katex] into the expression for the third interior angle.<\/p>\n\n\n\n [katex] \\begin{aligned}\\angle MPN &= \\angle (2x + 15)\\degree\\\\&= (2 \\cdot 25 +15)\\degree\\\\&= 65\\degree\\end{aligned} [\/katex]<\/p>\n\n\n\n Hence, [katex]\\angle MPN = 65\\degree[\/katex]. To check if we got the right angle measures, add the three interior angles again and see if they add up to 180o<\/sup>.<\/p>\n\n\n\n [katex]\\begin{aligned}\\angle MNP + \\angle NMP + \\angle MPN &= 180\\degree\\\\90\\degree +25\\degree +65 \\degree&= 180\\degree\\\\180\\degree &\\overset{\\checkmark}{=} 180\\degree \u200b\\end{aligned}[\/katex]<\/p>\n\n\n\n This confirms that the interior angles are correct. Apply a similar approach to solve more complex problems. <\/p>\n\n\n\n Some of the interior angles of the trapezoid, ABCD<\/em>, are as shown below. If [katex]\\angle BAD = \\angle ADC[\/katex], what is measure of [katex]\\angle BDC[\/katex]?<\/p>\n\n\n\n As can be seen from the diagram, the trapezoid is composed of two triangles: \u2206ADB and \u2206BDC. Each set of interior angles for these two triangles will share the same sum: 180o<\/sup>. In \u2206ADB, two of its interior angles\u2019 measures are already given, so find the measure of the third angle. Subtract the measures of the two angles from 180o<\/sup> to find the measure of [katex]\\angle ADB[\/katex].<\/p>\n\n\n\n [katex]\\begin{aligned}\\angle ADB &= 180\\degree – \\angle DAB \u2013 \\angle BAD\\\\&= 180\\degree \u2013 58\\degree \u2013 74\\degree\\\\&= 48\\degree\u200b\\end{aligned}[\/katex]<\/p>\n\n\n\n Now, we know that [katex]\\angle BAD = \\angle ADC[\/katex], so [katex]\\angle ADC = 58\\degree[\/katex]. To find the measure of [katex]\\angle BDC[\/katex], subtract the measure of [katex]\\angle ADB[\/katex] from that of [katex]\\angle ADC[\/katex].<\/p>\n\n\n\n [katex]\\begin{aligned}\\angle ADB + \\angle DBC &= \\angle ADC\\\\\\angle ADB + \\angle DBC&= 58\\degree\\\\48 \\degree + \\angle BDC &= 58\\degree\\\\\\angle BDC &= 10\\degree\u200b\\end{aligned}[\/katex]<\/p>\n\n\n\n Hence, [katex] \\angle BDC [\/katex] has a measure of [katex]10\\degree[\/katex].<\/p>\n","protected":false},"excerpt":{"rendered":" The triangle sum theorem establishes the common sum shared by the three interior triangles of any triangle. This article covers all the fundamentals needed to master this topic and find unknown angle measurements of triangles and polygons.<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"class_list":["post-472","post","type-post","status-publish","format-standard","hentry","category-geometry"],"_links":{"self":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/472","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/comments?post=472"}],"version-history":[{"count":1,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/472\/revisions"}],"predecessor-version":[{"id":481,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/posts\/472\/revisions\/481"}],"wp:attachment":[{"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/media?parent=472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/categories?post=472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/visualfractions.com\/blog\/wp-json\/wp\/v2\/tags?post=472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-7.png\")
<\/figure>\n\n\n\n\u2206ABC<\/strong><\/td> [katex]\\angle A + \\angle B + \\angle C = 60\\degree +50\\degree + 70\\degree =180\\degree[\/katex]<\/td><\/tr> \u2206MNP<\/strong><\/td> [katex]\\angle M + \\angle N + \\angle P = 40\\degree +90\\degree + 50\\degree =180\\degree[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Problem 1<\/h3>\n\n\n\n
![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-8.png\")
<\/figure>\n\n\n\n\u2206ABC<\/strong><\/td> [katex]\\angle A + \\angle B + \\angle C = 60\\degree +60\\degree + 60\\degree =180\\degree[\/katex]<\/td><\/tr> \u2206JKL<\/strong><\/td> [katex]\\angle J + \\angle K + \\angle L = 40\\degree +90\\degree + 50\\degree =180\\degree[\/katex]<\/td><\/tr> \u2206XYZ<\/strong><\/td> [katex]\\angle A + \\angle B + \\angle C = 40\\degree +90\\degree + 60\\degree =180\\degree[\/katex]<\/td><\/tr> \u2206MLN<\/strong><\/td> [katex]\\angle M + \\angle L + \\angle N = 50\\degree +70\\degree + 70\\degree =190\\degree[\/katex]<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Understanding the Triangle Sum Theorem\u2019s Proof<\/h2>\n\n\n\n
![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-9.png\")
<\/figure>\n\n\n\nHow To Solve Problems Using Triangle Sum Theorem?<\/h2>\n\n\n\n
Problem 2<\/h3>\n\n\n\n
![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-10.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-11.png\")
<\/figure>\n\n\n\nProblem 3<\/strong><\/h3>\n\n\n\n
![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-12.png\")
<\/figure>\n\n\n\nProblem 4<\/h3>\n\n\n\n
![\"\"](\"https:\/\/visualfractions.com\/blog\/wp-content\/uploads\/2022\/03\/image-13.png\")
<\/figure>\n\n\n\n