{"id":2125,"date":"2025-08-04T18:23:48","date_gmt":"2025-08-04T18:23:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2125"},"modified":"2025-10-17T21:28:29","modified_gmt":"2025-10-17T21:28:29","slug":"non-right-triangles-with-law-of-sines-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-sines-apply-it-1\/","title":{"raw":"Non-right Triangles with Law of Sines: Apply It 1","rendered":"Non-right Triangles with Law of Sines: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the Law of Sines to solve oblique triangles.<\/li>\r\n \t<li>Find the area of an oblique triangle using the sine function.<\/li>\r\n \t<li>Solve applied problems using the Law of Sines.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Applied Problems Using the Law of Sines<\/h2>\r\nThe more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165043\/CNX_Precalc_Figure_08_01_017.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.\" width=\"487\" height=\"134\" \/> <b>Figure 16<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"861681\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"861681\"]\r\n\r\nTo find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[\/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[\/latex].\r\n\r\nBecause the angles in the triangle add up to 180 degrees, the unknown angle must be 180\u00b0\u221215\u00b0\u221235\u00b0=130\u00b0. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\frac{\\sin \\left(130^\\circ \\right)}{20}=\\frac{\\sin \\left(35^\\circ \\right)}{a} \\\\ a\\sin \\left(130^\\circ \\right)=20\\sin \\left(35^\\circ \\right) \\\\ a=\\frac{20\\sin \\left(35^\\circ \\right)}{\\sin \\left(130^\\circ \\right)} \\\\ a\\approx 14.98 \\end{gathered}[\/latex]<\/p>\r\nThe distance from one station to the aircraft is about 14.98 miles.\r\n\r\nNow that we know [latex]a[\/latex], we can use right triangle relationships to solve for [latex]h[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(15^\\circ \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{a} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{14.98} \\\\ h=14.98\\sin \\left(15^\\circ \\right) \\\\ h\\approx 3.88 \\end{gathered}[\/latex]<\/p>\r\nThe aircraft is at an altitude of approximately 3.9 miles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\nThe diagram represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70\u00b0, the angle of elevation from the northern end zone, point [latex]B[\/latex], is 62\u00b0, and the distance between the viewing points of the two end zones is 145 yards.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165046\/CNX_Precalc_Figure_08_01_018.jpg\" alt=\"An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.\" width=\"487\" height=\"535\" \/>\r\n\r\n[reveal-answer q=\"826578\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"826578\"]\r\n\r\n161.9 yd\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]149308[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the Law of Sines to solve oblique triangles.<\/li>\n<li>Find the area of an oblique triangle using the sine function.<\/li>\n<li>Solve applied problems using the Law of Sines.<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Applied Problems Using the Law of Sines<\/h2>\n<p>The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165043\/CNX_Precalc_Figure_08_01_017.jpg\" alt=\"A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted altitude line perpendicular to the ground side connecting the airplane vertex with the ground.\" width=\"487\" height=\"134\" \/><figcaption class=\"wp-caption-text\"><b>Figure 16<\/b><\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q861681\">Show Solution<\/button><\/p>\n<div id=\"q861681\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side [latex]a[\/latex], and then use right triangle relationships to find the height of the aircraft, [latex]h[\/latex].<\/p>\n<p>Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180\u00b0\u221215\u00b0\u221235\u00b0=130\u00b0. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\frac{\\sin \\left(130^\\circ \\right)}{20}=\\frac{\\sin \\left(35^\\circ \\right)}{a} \\\\ a\\sin \\left(130^\\circ \\right)=20\\sin \\left(35^\\circ \\right) \\\\ a=\\frac{20\\sin \\left(35^\\circ \\right)}{\\sin \\left(130^\\circ \\right)} \\\\ a\\approx 14.98 \\end{gathered}[\/latex]<\/p>\n<p>The distance from one station to the aircraft is about 14.98 miles.<\/p>\n<p>Now that we know [latex]a[\/latex], we can use right triangle relationships to solve for [latex]h[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\left(15^\\circ \\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{a} \\\\ \\sin \\left(15^\\circ \\right)=\\frac{h}{14.98} \\\\ h=14.98\\sin \\left(15^\\circ \\right) \\\\ h\\approx 3.88 \\end{gathered}[\/latex]<\/p>\n<p>The aircraft is at an altitude of approximately 3.9 miles.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p>The diagram represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70\u00b0, the angle of elevation from the northern end zone, point [latex]B[\/latex], is 62\u00b0, and the distance between the viewing points of the two end zones is 145 yards.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27165046\/CNX_Precalc_Figure_08_01_018.jpg\" alt=\"An oblique triangle formed from three vertices A, B, and C. Verticies A and B are points on the ground, and vertex C is the blimp in the air between them. The distance between A and B is 145 yards. The angle at vertex A is 70 degrees, and the angle at vertex B is 62 degrees.\" width=\"487\" height=\"535\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q826578\">Show Solution<\/button><\/p>\n<div id=\"q826578\" class=\"hidden-answer\" style=\"display: none\">\n<p>161.9 yd<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm149308\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149308&theme=lumen&iframe_resize_id=ohm149308&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":221,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2125"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":2,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2125\/revisions"}],"predecessor-version":[{"id":4737,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2125\/revisions\/4737"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/221"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2125\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2125"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2125"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2125"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2125"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}