{"id":16698,"date":"2019-08-20T17:03:22","date_gmt":"2019-08-20T17:03:22","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/non-right-triangles-law-of-sines\/"},"modified":"2022-01-11T02:32:56","modified_gmt":"2022-01-11T02:32:56","slug":"solving-applied-problems-using-law-of-sines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/solving-applied-problems-using-law-of-sines\/","title":{"raw":"Solving Applied Problems using Law of Sines","rendered":"Solving Applied Problems using Law of Sines"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Solve applied problems using the Law of Sines.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165135381204\" class=\"bc-section section\">\r\n<p id=\"fs-id1165137628654\">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>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding an Altitude<\/h3>\r\n<p id=\"fs-id1165134041232\">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>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145153\/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\" \/> <strong>Figure 16.<\/strong>[\/caption]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1165132972747\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165132972747\"]\r\n<p id=\"fs-id1165132972749\">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>\r\n<p id=\"fs-id1165137732703\">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>\r\n\r\n<div id=\"fs-id1165134246237\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ \\text{ }\\frac{\\mathrm{sin}\\left(130\u00b0\\right)}{20}=\\frac{\\mathrm{sin}\\left(35\u00b0\\right)}{a}\\hfill \\end{array}\\hfill \\\\ a\\mathrm{sin}\\left(130\u00b0\\right)=20\\mathrm{sin}\\left(35\u00b0\\right)\\hfill \\\\ \\text{ }a=\\frac{20\\mathrm{sin}\\left(35\u00b0\\right)}{\\mathrm{sin}\\left(130\u00b0\\right)}\\hfill \\\\ \\text{ }a\\approx 14.98\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165133102541\">The distance from one station to the aircraft is about 14.98 miles.<\/p>\r\n<p id=\"fs-id1165133102544\">Now that we know[latex],a,,[\/latex]we can use right triangle relationships to solve for[latex],h.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135172284\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\mathrm{sin}\\left(15\u00b0\\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}}\\hfill \\\\ \\mathrm{sin}\\left(15\u00b0\\right)=frac{h}{a}\\hfill \\\\ \\mathrm{sin}\\left(15\u00b0\\right)=\\frac{h}{14.98}\\hfill \\\\ \\text{ }\\text{ }h=14.98\\mathrm{sin}\\left(15\u00b0\\right)\\hfill \\\\ \\text{ }h\\approx 3.88\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137722150\">The aircraft is at an altitude of approximately 3.9 miles.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\nThe diagram shown in (Figure 17) 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[caption id=\"\" align=\"aligncenter\" width=\"487\" class=\"small\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145155\/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\" \/> <strong>Figure 17.<\/strong>[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1165134380826\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134380826\"]\r\n<p id=\"fs-id1165134380829\">161.9 yd.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165135646148\">\r\n \t<dt>altitude<\/dt>\r\n \t<dd id=\"fs-id1165135646153\">a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134357482\">\r\n \t<dt>ambiguous case<\/dt>\r\n \t<dd id=\"fs-id1165134357487\">a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134357493\">\r\n \t<dt>Law of Sines<\/dt>\r\n \t<dd id=\"fs-id1165133234509\">states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165133234516\">\r\n \t<dt>oblique triangle<\/dt>\r\n \t<dd id=\"fs-id1165135551173\">any triangle that is not a right triangle<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Solve applied problems using the Law of Sines.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135381204\" class=\"bc-section section\">\n<p id=\"fs-id1165137628654\">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<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding an Altitude<\/h3>\n<p id=\"fs-id1165134041232\">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<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145153\/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\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 16.<\/strong><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165132972747\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165132972747\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165132972749\">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 id=\"fs-id1165137732703\">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<div id=\"fs-id1165134246237\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\end{array}\\hfill \\\\ \\text{ }\\frac{\\mathrm{sin}\\left(130\u00b0\\right)}{20}=\\frac{\\mathrm{sin}\\left(35\u00b0\\right)}{a}\\hfill \\end{array}\\hfill \\\\ a\\mathrm{sin}\\left(130\u00b0\\right)=20\\mathrm{sin}\\left(35\u00b0\\right)\\hfill \\\\ \\text{ }a=\\frac{20\\mathrm{sin}\\left(35\u00b0\\right)}{\\mathrm{sin}\\left(130\u00b0\\right)}\\hfill \\\\ \\text{ }a\\approx 14.98\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165133102541\">The distance from one station to the aircraft is about 14.98 miles.<\/p>\n<p id=\"fs-id1165133102544\">Now that we know[latex],a,,[\/latex]we can use right triangle relationships to solve for[latex],h.[\/latex]<\/p>\n<div id=\"fs-id1165135172284\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{l}\\mathrm{sin}\\left(15\u00b0\\right)=\\frac{\\text{opposite}}{\\text{hypotenuse}}\\hfill \\\\ \\mathrm{sin}\\left(15\u00b0\\right)=frac{h}{a}\\hfill \\\\ \\mathrm{sin}\\left(15\u00b0\\right)=\\frac{h}{14.98}\\hfill \\\\ \\text{ }\\text{ }h=14.98\\mathrm{sin}\\left(15\u00b0\\right)\\hfill \\\\ \\text{ }h\\approx 3.88\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137722150\">The aircraft is at an altitude of approximately 3.9 miles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p>The diagram shown in (Figure 17) 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<div style=\"width: 497px\" class=\"wp-caption aligncenter small\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19145155\/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<p class=\"wp-caption-text\"><strong>Figure 17.<\/strong><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134380826\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134380826\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134380829\">161.9 yd.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135646148\">\n<dt>altitude<\/dt>\n<dd id=\"fs-id1165135646153\">a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134357482\">\n<dt>ambiguous case<\/dt>\n<dd id=\"fs-id1165134357487\">a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134357493\">\n<dt>Law of Sines<\/dt>\n<dd id=\"fs-id1165133234509\">states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side<\/dd>\n<\/dl>\n<dl id=\"fs-id1165133234516\">\n<dt>oblique triangle<\/dt>\n<dd id=\"fs-id1165135551173\">any triangle that is not a right triangle<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":473810,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["jay-abramson"],"pb_section_license":"cc-by"},"chapter-type":[],"contributor":[75],"license":[57],"class_list":["post-16698","chapter","type-chapter","status-publish","hentry","contributor-jay-abramson","license-cc-by"],"part":14036,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16698","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/users\/473810"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16698\/revisions"}],"predecessor-version":[{"id":16716,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16698\/revisions\/16716"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/parts\/14036"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapters\/16698\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/media?parent=16698"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/pressbooks\/v2\/chapter-type?post=16698"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/contributor?post=16698"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/wp-json\/wp\/v2\/license?post=16698"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}