{"id":241,"date":"2017-08-23T10:29:13","date_gmt":"2017-08-23T10:29:13","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/mathematics\/?page_id=241"},"modified":"2017-09-22T16:07:46","modified_gmt":"2017-09-22T16:07:46","slug":"law-of-sines-law-of-cosines","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/mathematics\/law-of-sines-law-of-cosines\/","title":{"rendered":"Law of Sines &#038; Law of Cosines"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometric-ratios\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometry\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/the-unit-circle\">Next Lesson \u27a1<\/a><\/div>\n<p><!-- CONTENT BEGINS HERE --><\/p>\n<h1 id=\"title\">Law of Sines &amp; Law of Cosines<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson, you will learn how to apply the Law of Sines and the Law of Cosines to determine the length of sides\u00a0and measure of angles for triangles other than right triangles.<\/p>\n<h4>Previously Covered:<\/h4>\n<ul>\n<li>The\u00a0area of a triangle is equal to one-half times its base times its\u00a0height, or\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p1_clip_image008.gif\" width=\"33\" height=\"41\" name=\"graphics2\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li><strong>Sine\u00a0<\/strong>is equal to the measure of the opposite leg over the length of\u00a0the hypotenuse, or\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p1_clip_image011.gif\" width=\"127\" height=\"44\" name=\"graphics3\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li><strong>Cosine\u00a0<\/strong>is equal to length of adjacent leg over length of hypotenuse,\u00a0expressed as a formula as\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p1_clip_image014.gif\" width=\"129\" height=\"44\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/li>\n<li>The sum of the interior angles\u00a0of a triangle is always equal to 180\u00b0.<\/li>\n<li>All sides of an equilateral triangle are equal in length.\u00a0The interior angle measurements are 60\u00b0 for each angle in an\u00a0equilateral triangle.<\/li>\n<\/ul>\n<section>\n<h3>What are the Law of Sines and the Law of Cosines?<\/h3>\n<p class=\"lesson_text\">The <abbr title=\"The ratio of two sides is equal to the ratio of the sines of their opposite angles: c. \">Law of\u00a0Sines<\/abbr> and the <abbr title=\"a law that tells us how to solve for side c when we know the lengths of sides a and b, and we also know the measurement of the angle between sides a and b: c. \">Law\u00a0of Cosines<\/abbr> are used to determine the length of sides or\u00a0measure of angles of triangles that are not right triangles.<\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">Let <\/span><em><span style=\"text-decoration: none;\">A,\u00a0B<\/span><\/em><span style=\"text-decoration: none;\">, and <\/span><em><span style=\"text-decoration: none;\">C\u00a0<\/span><\/em><span style=\"text-decoration: none;\">represent the three interior angles of any triangle. Let <\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">denote the length of the side that is opposite the angle <\/span><em><span style=\"text-decoration: none;\">A<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0<\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">denote the length of the side that is opposite the angle <\/span><em><span style=\"text-decoration: none;\">B<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and <\/span><em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">denote the length of the side that is opposite the angle <\/span><em><span style=\"text-decoration: none;\">C<\/span><\/em><span style=\"text-decoration: none;\">.<\/span><\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">The La<\/span>w of Sines\u00a0states that the ratio of two sides is equal to the ratio of the\u00a0sines of their opposite angles. So the formula for the Law of\u00a0Sines is<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p2_clip_image003.gif\" width=\"165\" height=\"41\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20009.JPG\" alt=\"law of sines\" width=\"198\" height=\"164\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">The Law of Sines can be used to solve for all of the angles of\u00a0a triangle if all of the sides of the triangle are known. The Law\u00a0of Sines can also be used to help solve for lengths of sides and\u00a0measurements of angles. If you know two sides and one angle, you\u00a0can find the length of the third side.<\/p>\n<p class=\"lesson_text\">Make sure you pay attention to the definition and formula for\u00a0the Law of Sines because you will be asked prove the Law of Sines\u00a0at the end of this module.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p class=\"lesson_text\">What is the length of side <em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em>in the triangle on the right?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20010.JPG\" alt=\"triangle with sides, a and 5 m, and angles 35 and 65 degrees\" width=\"232\" height=\"130\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>2.7 meters<\/li>\n<li>3.9 meters<\/li>\n<li>2.3 meters<\/li>\n<li>5.2 meters<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p class=\"lesson_text\">The correct choice is B. If we use the Law of Sines, we know\u00a0that<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p2_clip_image006.gif\" width=\"117\" height=\"41\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">By using cross-multiplication, we can also express the Law of\u00a0Sines as<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p2_clip_image009.gif\" width=\"129\" height=\"21\" name=\"graphics7\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">Therefore,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p2_clip_image012.gif\" width=\"87\" height=\"44\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">So, the correct answer is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p2_clip_image015.gif\" width=\"61\" height=\"19\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<\/div>\n<\/section>\n<h3>What is the Law of Cosines?<\/h3>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">The Law of Cosines is used\u00a0when we know the lengths of sides <\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and we also know the measurement of the angle between sides <\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0which as you can see in the triangle below, is angle <\/span><em><span style=\"text-decoration: none;\">C<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0Look at the triangle with sides <\/span><em><span style=\"text-decoration: none;\">a<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0<\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0The side <\/span><em><span style=\"text-decoration: none;\">h\u00a0<\/span><\/em><span style=\"text-decoration: none;\">is perpendicular to side <\/span><em><span style=\"text-decoration: none;\">a,\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and is included to split the triangle into two right triangles.\u00a0Remember that splitting a triangle that is not a right triangle\u00a0into more than one right triangle (through the use of a line that\u00a0is perpendicular to a side of the existing triangle) is a tool\u00a0that you have that will enable you to apply trigonometric formulas\u00a0in cases where, at first glance, you may have thought you could\u00a0not apply them. <\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20011.JPG\" alt=\"Law of cosines\" width=\"278\" height=\"130\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">The Law of Cosines can be shown as a formula by\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image003.gif\" width=\"161\" height=\"21\" name=\"graphics4\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<h4>Proof of the Law of Cosines<\/h4>\n<p class=\"lesson_text\">Using the Pythagorean Theorem, we know\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image006.gif\" width=\"117\" height=\"25\" name=\"graphics5\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0which is expressed in expanded form as\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image009.gif\" width=\"156\" height=\"21\" name=\"graphics6\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">Also using the Pythagorean Theorem, we know that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image012.gif\" width=\"85\" height=\"21\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">By substituting\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image015.gif\" width=\"19\" height=\"21\" name=\"graphics8\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0for\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image018.gif\" width=\"52\" height=\"21\" name=\"graphics9\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0we get\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image021.gif\" width=\"123\" height=\"21\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">At this point, we have\u00a0solved for sides <\/span><em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and for angle <\/span><em><span style=\"text-decoration: none;\">C<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0We cannot stop here because the variable <\/span><em><span style=\"text-decoration: none;\">d\u00a0<\/span><\/em><span style=\"text-decoration: none;\">must be solved for in order to finish the equation. Since we can\u00a0choose to look at only one of the right triangles, we can apply\u00a0one of our trigonometric ratios that we learned earlier. Do you\u00a0remember <\/span><em><span style=\"text-decoration: none;\">soh cah\u00a0toa<\/span><\/em><span style=\"text-decoration: none;\">? Since we are\u00a0trying to find an equation for <\/span><em><span style=\"text-decoration: none;\">d<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0we know that we need to look at a trigonometric ratio that\u00a0includes an <\/span>adjacent side. Therefore, we are looking at\u00a0either the \u201cC-A-H\u201d to get\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image024_0000.gif\" width=\"129\" height=\"44\" name=\"graphics11\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0or the \u201cT-O-A\u201d to get\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image027.gif\" width=\"112\" height=\"44\" name=\"graphics12\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0<span style=\"text-decoration: none;\">Now we have sides, <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">and <\/span><em><span style=\"text-decoration: none;\">h\u00a0<\/span><\/em><span style=\"text-decoration: none;\">in the triangle to look at. Since we do not know side <\/span><em><span style=\"text-decoration: none;\">h,\u00a0<\/span><\/em><span style=\"text-decoration: none;\">we should look at side <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0Side <\/span><em><span style=\"text-decoration: none;\">b <\/span><\/em><span style=\"text-decoration: none;\">is\u00a0the hypotenuse in the right triangle with sides <\/span><em><span style=\"text-decoration: none;\">b<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0<\/span><em><span style=\"text-decoration: none;\">d<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0and <\/span><em><span style=\"text-decoration: none;\">h<\/span><\/em><span style=\"text-decoration: none;\">.\u00a0So we know to use the trigonometric ratio\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image024_0000.gif\" width=\"129\" height=\"44\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/><\/span>.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image031.gif\" width=\"133\" height=\"89\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">T<span style=\"text-decoration: none;\">herefore, since we are\u00a0attempting to solve for <\/span><em><span style=\"text-decoration: none;\">d,\u00a0<\/span><\/em><span style=\"text-decoration: none;\">we need to multiply both sides by the hypotenuse <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">to get\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image034.gif\" width=\"76\" height=\"19\" name=\"graphics15\" align=\"ABSMIDDLE\" border=\"0\" \/>.<br \/>\n<\/span><\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">Now we can plug in\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image037.gif\" width=\"49\" height=\"19\" name=\"graphics16\" align=\"ABSMIDDLE\" border=\"0\" \/>\u00a0for <\/span><em><span style=\"text-decoration: none;\">d\u00a0<\/span><\/em><span style=\"text-decoration: none;\">in the equation\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image021_0000.gif\" width=\"123\" height=\"21\" name=\"graphics17\" align=\"ABSMIDDLE\" border=\"0\" \/>.\u00a0This gives us\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image003_0000.gif\" width=\"161\" height=\"21\" name=\"graphics18\" align=\"ABSMIDDLE\" border=\"0\" \/>.<br \/>\n<\/span><\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">This proves the Law of\u00a0Cosines,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p3_clip_image003_0001.gif\" width=\"161\" height=\"21\" name=\"graphics19\" align=\"ABSMIDDLE\" border=\"0\" \/><\/span>.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p class=\"lesson_text\">Using the Law of Cosines and the triangle on the right, what is\u00a0the value <span style=\"text-decoration: none;\">of <\/span><em><span style=\"text-decoration: none;\">c<\/span><\/em><span style=\"text-decoration: none;\">?<\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20012.JPG\" alt=\"equilateral triangle with sides 10 m\" width=\"164\" height=\"162\" name=\"graphics3\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>9 meters<\/li>\n<li>10 meters<\/li>\n<li>11 meters<\/li>\n<li>15 meters<\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p class=\"lesson_text\">The correct choice is B. Using the Law of Cosines, we know\u00a0that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image003.gif\" width=\"165\" height=\"21\" name=\"graphics4\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">Since <em><span style=\"text-decoration: none;\">a\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 10, <\/span><em><span style=\"text-decoration: none;\">b\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 10, and <\/span><em><span style=\"text-decoration: none;\">C\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 60 \u00b0 , <\/span>we can use substitution.<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image006.gif\" width=\"240\" height=\"71\" name=\"graphics5\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">Y<span style=\"text-decoration: none;\">ou can simply surmise by\u00a0looking at the triangle that <\/span><em><span style=\"text-decoration: none;\">c\u00a0<\/span><\/em><span style=\"text-decoration: none;\">= 10 m, because the triangle above is an equilateral triangle.\u00a0Equilateral triangles, as we have learned, have all sides of\u00a0equal length and all equal angle measurements (angles in an\u00a0equilateral triangle are always 60\u00b0 acute angles). The Law of\u00a0cosines may not have been the ea<\/span>siest way to solve this\u00a0particular problem, but it is nonetheless important to completely\u00a0understand the concept.<\/p>\n<\/div>\n<\/section>\n<p class=\"lesson_text\">Since you have tried an example that required use of the law of\u00a0sines and an example that illustrated the law of cosines, we will\u00a0try an example that uses both the law of sines and the law of\u00a0cosines.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<div>\n<p class=\"lesson_text\">What is the area of the triangle below?<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20013.JPG\" alt=\"Triangle with angle 20\" width=\"289\" height=\"165\" name=\"graphics6\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<ol>\n<li>24.98 m<sup>2<\/sup><\/li>\n<li>45.26 m<sup>2<\/sup><\/li>\n<li>58.18 m<sup>2<\/sup><\/li>\n<li><a name=\"two\"><\/a>82 m<sup>2<\/sup><\/li>\n<\/ol>\n<\/div>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p class=\"lesson_text\">The correct choice is C. We know from the Law of Sines that\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image009.gif\" width=\"91\" height=\"41\" name=\"graphics7\" align=\"ABSMIDDLE\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">First, we must solve for each missing variable.<\/p>\n<p align=\"CENTER\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image012.gif\" width=\"132\" height=\"113\" name=\"graphics8\" align=\"BOTTOM\" border=\"0\" \/><\/strong><\/p>\n<p class=\"lesson_text\"><span style=\"text-decoration: none;\">Remember the rule that\u00a0says, in order to solve for <\/span><em><span style=\"text-decoration: none;\">A<\/span><\/em><span style=\"text-decoration: none;\">,\u00a0you can say that<\/span><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image015.gif\" width=\"136\" height=\"51\" name=\"graphics9\" align=\"BOTTOM\" border=\"0\" \/>.<\/p>\n<p class=\"lesson_text\">Therefore,\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image018.gif\" width=\"75\" height=\"21\" name=\"graphics10\" align=\"ABSMIDDLE\" border=\"0\" \/>,\u00a0and since the interior angles of a triangle adds up to 180\u00b0,<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/s2_p4_html_m7431f5ec.gif\" width=\"217\" height=\"21\" name=\"graphics18\" align=\"BOTTOM\" border=\"0\" \/>.<br \/>\nUsing the Law of Cosines, we can say that<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image021.gif\" width=\"261\" height=\"71\" name=\"graphics11\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/Math%20Mod%206.1%20Art%20014.JPG\" alt=\"Triangle Rotated\" width=\"288\" height=\"157\" name=\"graphics12\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p class=\"lesson_text\">It is easiest to rotate the triangle as shown above.<\/p>\n<p class=\"lesson_text\">The ar<span style=\"text-decoration: none;\">ea of a triangle is\u00a0<\/span><strong><span style=\"text-decoration: none;\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image024.gif\" width=\"135\" height=\"41\" name=\"graphics13\" align=\"ABSMIDDLE\" border=\"0\" \/>.<br \/>\n<\/span><\/strong><\/p>\n<p class=\"lesson_text\">Since we know the base, we only need to solve for the height\u00a0in order to find the area.<\/p>\n<p class=\"lesson_text\">Using the Law of Sines, we can determine that<\/p>\n<p align=\"CENTER\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image027.gif\" width=\"161\" height=\"135\" name=\"graphics14\" align=\"BOTTOM\" border=\"0\" \/><\/p>\n<p align=\"CENTER\"><strong><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/images\/s2_p4_clip_image030.gif\" width=\"173\" height=\"108\" name=\"graphics15\" align=\"BOTTOM\" border=\"0\" \/><\/strong><\/p>\n<p class=\"lesson_text\">Therefore, the area of the triangle is\u00a0<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/math\/6\/s2_p4_html_1ba6fa8f.gif\" width=\"57\" height=\"21\" name=\"graphics19\" align=\"absmiddle\" border=\"0\" \/>.<\/p>\n<\/div>\n<\/section>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometric-ratios\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/trigonometry\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/mathematics\/the-unit-circle\">Next Lesson \u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Law of Sines &amp; Law of Cosines Objective In this lesson, you will learn how to apply the Law of Sines and the Law of Cosines to determine the length of sides\u00a0and measure of angles for triangles other than right triangles. Previously Covered: The\u00a0area of a triangle is equal [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-241","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/241","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/comments?post=241"}],"version-history":[{"count":9,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/241\/revisions"}],"predecessor-version":[{"id":799,"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/pages\/241\/revisions\/799"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/mathematics\/wp-json\/wp\/v2\/media?parent=241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}