{"id":1561,"date":"2025-07-25T03:41:56","date_gmt":"2025-07-25T03:41:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1561"},"modified":"2026-03-12T06:46:41","modified_gmt":"2026-03-12T06:46:41","slug":"non-right-triangles-with-law-of-sines-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-sines-fresh-take\/","title":{"raw":"Non-right Triangles with Law of Sines: Fresh Take","rendered":"Non-right Triangles with Law of Sines: Fresh Take"},"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 Oblique Triangles with the Law of Sines<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<p data-start=\"76\" data-end=\"520\">The Law of Sines is a powerful tool for solving <strong data-start=\"124\" data-end=\"145\">oblique triangles<\/strong> (non-right triangles). It relates the ratios of side lengths to the sines of their opposite angles, allowing us to find missing sides or angles when certain combinations of information are known. The formula works for both acute and obtuse triangles, but care must be taken with the <strong data-start=\"429\" data-end=\"453\">ambiguous case (SSA)<\/strong>, where two different triangles may satisfy the given conditions.<\/p>\r\n<p data-start=\"522\" data-end=\"541\">The Law of Sines:<\/p>\r\n<p data-start=\"543\" data-end=\"617\">[latex]\\dfrac{a}{\\sin A} = \\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}[\/latex]<\/p>\r\n<strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Using the Law of Sines<\/strong>\r\n<ol>\r\n \t<li data-start=\"666\" data-end=\"864\">\r\n<p data-start=\"669\" data-end=\"694\"><strong data-start=\"669\" data-end=\"692\">Know When to Use It<\/strong><\/p>\r\n\r\n<ul data-start=\"698\" data-end=\"864\">\r\n \t<li data-start=\"698\" data-end=\"776\">\r\n<p data-start=\"700\" data-end=\"776\">Works best with <strong data-start=\"716\" data-end=\"723\">ASA<\/strong> (angle-side-angle) and <strong data-start=\"747\" data-end=\"754\">AAS<\/strong> (angle-angle-side).<\/p>\r\n<\/li>\r\n \t<li data-start=\"780\" data-end=\"864\">\r\n<p data-start=\"782\" data-end=\"864\">Also used in <strong data-start=\"795\" data-end=\"802\">SSA<\/strong> (side-side-angle), but this may lead to the ambiguous case.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li data-start=\"866\" data-end=\"1028\">\r\n<p data-start=\"869\" data-end=\"888\"><strong data-start=\"869\" data-end=\"886\">Set Up Ratios<\/strong><\/p>\r\n\r\n<ul data-start=\"892\" data-end=\"1028\">\r\n \t<li data-start=\"892\" data-end=\"973\">\r\n<p data-start=\"894\" data-end=\"973\">Match a known side with its opposite angle: [latex]\\dfrac{a}{\\sin A}[\/latex].<\/p>\r\n<\/li>\r\n \t<li data-start=\"977\" data-end=\"1028\">\r\n<p data-start=\"979\" data-end=\"1028\">Use that ratio to find missing sides or angles.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li data-start=\"1030\" data-end=\"1231\">\r\n<p data-start=\"1033\" data-end=\"1063\"><strong data-start=\"1033\" data-end=\"1061\">Solve for Missing Angles<\/strong><\/p>\r\n\r\n<ul data-start=\"1067\" data-end=\"1231\">\r\n \t<li data-start=\"1067\" data-end=\"1149\">\r\n<p data-start=\"1069\" data-end=\"1149\">Use inverse sine: [latex]B = \\sin^{-1}\\left(\\dfrac{b\\sin A}{a}\\right)[\/latex].<\/p>\r\n<\/li>\r\n \t<li data-start=\"1153\" data-end=\"1231\">\r\n<p data-start=\"1155\" data-end=\"1231\">Be mindful of possible supplementary solutions (an acute or obtuse angle).<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li data-start=\"1233\" data-end=\"1392\">\r\n<p data-start=\"1236\" data-end=\"1261\"><strong data-start=\"1236\" data-end=\"1259\">Finish the Triangle<\/strong><\/p>\r\n\r\n<ul data-start=\"1265\" data-end=\"1392\">\r\n \t<li data-start=\"1265\" data-end=\"1392\">\r\n<p data-start=\"1267\" data-end=\"1392\">After finding two angles, the third can be calculated by subtracting from [latex]180^\\circ[\/latex] (or [latex]\\pi[\/latex]).<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>\r\n<p data-start=\"1267\" data-end=\"1392\"><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\" data-start=\"1747\" data-end=\"1770\">Check for Ambiguity<\/strong><\/p>\r\n\r\n<ul data-start=\"1776\" data-end=\"1898\">\r\n \t<li data-start=\"1776\" data-end=\"1898\">\r\n<p data-start=\"1778\" data-end=\"1898\">In SSA cases, if [latex]\\sin^{-1}[\/latex] gives an acute angle, check if its supplement also creates a valid triangle.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div><section class=\"textbox example\">\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">In triangle ABC, [latex]A = 35\u00b0[\/latex], [latex]C = 65\u00b0[\/latex], and [latex]c = 10[\/latex] cm. Find side [latex]a[\/latex] and angle [latex]B[\/latex].<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[reveal-answer q=\"948825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"948825\"]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">First, find angle [latex]B[\/latex]:<\/p>\r\n[latex] \\begin{aligned} B &amp;= 180\u00b0 - A - C \\\\ B &amp;= 180\u00b0 - 35\u00b0 - 65\u00b0 \\\\ B &amp;= 80\u00b0 \\end{aligned} [\/latex]\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">Now use the Law of Sines to find side [latex]a[\/latex]:<\/p>\r\n[latex] \\begin{aligned} \\frac{a}{\\sin A} &amp;= \\frac{c}{\\sin C} \\\\ \\frac{a}{\\sin 35\u00b0} &amp;= \\frac{10}{\\sin 65\u00b0} \\\\ a &amp;= \\frac{10 \\sin 35\u00b0}{\\sin 65\u00b0} \\\\ a &amp;= \\frac{10(0.5736)}{0.9063} \\\\ a &amp;\\approx 6.33 \\text{ cm} \\end{aligned} [\/latex]\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Therefore, [latex]B = 80\u00b0[\/latex] and [latex]a \\approx 6.33[\/latex] cm.<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><\/div>\r\n<div><section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-gfdeddhc-RCyjglaJo5w\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/RCyjglaJo5w?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-gfdeddhc-RCyjglaJo5w\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14661402&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-gfdeddhc-RCyjglaJo5w&vembed=0&video_id=RCyjglaJo5w&video_target=tpm-plugin-gfdeddhc-RCyjglaJo5w'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/The+Ambiguous+Case+for+Sine+Law+-+Nerdstudy_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cThe Ambiguous Case for Sine Law - Nerdstudy\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><\/div>\r\n<h2>Area of an Oblique Triangle with Sine<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<p data-start=\"66\" data-end=\"360\">When a triangle isn\u2019t a right triangle, we can still find its area using the sine function. Instead of needing the height, we use two sides and the sine of the included angle. This is especially useful for <strong data-start=\"272\" data-end=\"279\">SAS<\/strong> (side\u2013angle\u2013side) cases, where two sides and the angle between them are known.<\/p>\r\n<p data-start=\"362\" data-end=\"379\">The formula is:<\/p>\r\n<p data-start=\"381\" data-end=\"478\">[latex]\\text{Area} = \\dfrac{1}{2}ab\\sin C = \\dfrac{1}{2}bc\\sin A = \\dfrac{1}{2}ca\\sin B[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div><section class=\"textbox example\">\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Find the area of triangle ABC where [latex]a = 8[\/latex] m, [latex]c = 14[\/latex] m, and [latex]B = 55\u00b0[\/latex].<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[reveal-answer q=\"566485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"566485\"]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">We have two sides and the included angle between them, so use:<\/p>\r\n[latex] \\text{Area} = \\frac{1}{2}ac\\sin B [\/latex]\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">Substitute the values:<\/p>\r\n[latex] \\begin{aligned} \\text{Area} &amp;= \\frac{1}{2}(8)(14)\\sin 55\u00b0 \\\\ &amp;= \\frac{1}{2}(112)\\sin 55\u00b0 \\\\ &amp;= 56 \\sin 55\u00b0 \\\\ &amp;= 56(0.8192) \\\\ &amp;\\approx 45.9 \\text{ m}^2 \\end{aligned} [\/latex]\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">The area of the triangle is approximately [latex]45.9[\/latex] square meters.<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><\/div>\r\n<div><section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-chefaaha-mBFDq4bPXMs\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/mBFDq4bPXMs?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-chefaaha-mBFDq4bPXMs\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14661403&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-chefaaha-mBFDq4bPXMs&vembed=0&video_id=mBFDq4bPXMs&video_target=tpm-plugin-chefaaha-mBFDq4bPXMs'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/The+Area+of+a+Triangle+using+Sine_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cThe Area of a Triangle using Sine\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><\/div>\r\n<h2>Applied Problems with the Law of Sines<\/h2>\r\n<div><section class=\"textbox example\">\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Two ranger stations are 15 miles apart, with station A directly east of station B. A fire is spotted from both stations. From station A, the angle to the fire is [latex]42\u00b0[\/latex] west of north. From station B, the angle to the fire is [latex]64\u00b0[\/latex] east of north. How far is the fire from station A?<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[reveal-answer q=\"76468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"76468\"]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Draw a diagram with:<\/p>\r\n\r\n<ul class=\"[li_&amp;]:mb-0 [li_&amp;]:mt-1 [li_&amp;]:gap-1 [&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc flex flex-col gap-1 pl-8 mb-3\">\r\n \t<li class=\"whitespace-normal break-words pl-2\">Stations A and B are 15 miles apart<\/li>\r\n \t<li class=\"whitespace-normal break-words pl-2\">The fire forms a triangle with the two stations<\/li>\r\n<\/ul>\r\n<img class=\"alignnone wp-image-5630\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM.png\" alt=\"The image shows a triangle formed by three points labeled A, B, and Fire. Points A and B lie horizontally along the bottom and are 15 miles apart.\" width=\"344\" height=\"216\" \/>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">The angle at station A (interior angle of triangle) is: [latex]90\u00b0 - 42\u00b0 = 48\u00b0[\/latex]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">The angle at station B (interior angle of triangle) is: [latex]90\u00b0 - 64\u00b0 = 26\u00b0[\/latex]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">The angle at the fire location is: [latex]180\u00b0 - 48\u00b0 - 26\u00b0 = 106\u00b0[\/latex]<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Let [latex]d[\/latex] be the distance from station A to the fire. The side opposite station A (from B to fire) is what we need, and the side opposite the fire (from A to B) is 15 miles.<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-[1.7]\">Using the Law of Sines:<\/p>\r\n[latex] \\begin{aligned} \\frac{15}{\\sin 106\u00b0} &amp;= \\frac{d}{\\sin 26\u00b0} \\\\ d &amp;= \\frac{15 \\sin 26\u00b0}{\\sin 106\u00b0} \\\\ d &amp;= \\frac{15(0.4384)}{0.9613} \\\\ d &amp;\\approx 6.8 \\text{ miles} \\end{aligned} [\/latex]\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">The fire is approximately [latex]6.8[\/latex] miles from station A.<\/p>\r\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">[\/hidden-answer]<\/p>\r\n\r\n<\/section><\/div>\r\n<div><section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api \"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-gcfbabcb-6dEvIdTJ708\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/6dEvIdTJ708?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-gcfbabcb-6dEvIdTJ708\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script type='text\/javascript' src='\/\/plugin.3playmedia.com\/ajax.js?cc=1&cc_minimizable=1&cc_minimize_on_load=0&cc_multi_text_track=0&cc_overlay=1&cc_searchable=0&embed=ajax&mf=14661404&p3sdk_version=1.11.7&p=20361&player_type=youtube&plugin_skin=dark&target=3p-plugin-target-gcfbabcb-6dEvIdTJ708&vembed=0&video_id=6dEvIdTJ708&video_target=tpm-plugin-gcfbabcb-6dEvIdTJ708'><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Example+-+Application+Problem+Solved+Using+the+Law+of+Sines_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Application Problem Solved Using the Law of Sines\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><\/div>","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 Oblique Triangles with the Law of Sines<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p data-start=\"76\" data-end=\"520\">The Law of Sines is a powerful tool for solving <strong data-start=\"124\" data-end=\"145\">oblique triangles<\/strong> (non-right triangles). It relates the ratios of side lengths to the sines of their opposite angles, allowing us to find missing sides or angles when certain combinations of information are known. The formula works for both acute and obtuse triangles, but care must be taken with the <strong data-start=\"429\" data-end=\"453\">ambiguous case (SSA)<\/strong>, where two different triangles may satisfy the given conditions.<\/p>\n<p data-start=\"522\" data-end=\"541\">The Law of Sines:<\/p>\n<p data-start=\"543\" data-end=\"617\">[latex]\\dfrac{a}{\\sin A} = \\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}[\/latex]<\/p>\n<p><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Quick Tips: Using the Law of Sines<\/strong><\/p>\n<ol>\n<li data-start=\"666\" data-end=\"864\">\n<p data-start=\"669\" data-end=\"694\"><strong data-start=\"669\" data-end=\"692\">Know When to Use It<\/strong><\/p>\n<ul data-start=\"698\" data-end=\"864\">\n<li data-start=\"698\" data-end=\"776\">\n<p data-start=\"700\" data-end=\"776\">Works best with <strong data-start=\"716\" data-end=\"723\">ASA<\/strong> (angle-side-angle) and <strong data-start=\"747\" data-end=\"754\">AAS<\/strong> (angle-angle-side).<\/p>\n<\/li>\n<li data-start=\"780\" data-end=\"864\">\n<p data-start=\"782\" data-end=\"864\">Also used in <strong data-start=\"795\" data-end=\"802\">SSA<\/strong> (side-side-angle), but this may lead to the ambiguous case.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<li data-start=\"866\" data-end=\"1028\">\n<p data-start=\"869\" data-end=\"888\"><strong data-start=\"869\" data-end=\"886\">Set Up Ratios<\/strong><\/p>\n<ul data-start=\"892\" data-end=\"1028\">\n<li data-start=\"892\" data-end=\"973\">\n<p data-start=\"894\" data-end=\"973\">Match a known side with its opposite angle: [latex]\\dfrac{a}{\\sin A}[\/latex].<\/p>\n<\/li>\n<li data-start=\"977\" data-end=\"1028\">\n<p data-start=\"979\" data-end=\"1028\">Use that ratio to find missing sides or angles.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<li data-start=\"1030\" data-end=\"1231\">\n<p data-start=\"1033\" data-end=\"1063\"><strong data-start=\"1033\" data-end=\"1061\">Solve for Missing Angles<\/strong><\/p>\n<ul data-start=\"1067\" data-end=\"1231\">\n<li data-start=\"1067\" data-end=\"1149\">\n<p data-start=\"1069\" data-end=\"1149\">Use inverse sine: [latex]B = \\sin^{-1}\\left(\\dfrac{b\\sin A}{a}\\right)[\/latex].<\/p>\n<\/li>\n<li data-start=\"1153\" data-end=\"1231\">\n<p data-start=\"1155\" data-end=\"1231\">Be mindful of possible supplementary solutions (an acute or obtuse angle).<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<li data-start=\"1233\" data-end=\"1392\">\n<p data-start=\"1236\" data-end=\"1261\"><strong data-start=\"1236\" data-end=\"1259\">Finish the Triangle<\/strong><\/p>\n<ul data-start=\"1265\" data-end=\"1392\">\n<li data-start=\"1265\" data-end=\"1392\">\n<p data-start=\"1267\" data-end=\"1392\">After finding two angles, the third can be calculated by subtracting from [latex]180^\\circ[\/latex] (or [latex]\\pi[\/latex]).<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<li>\n<p data-start=\"1267\" data-end=\"1392\"><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\" data-start=\"1747\" data-end=\"1770\">Check for Ambiguity<\/strong><\/p>\n<ul data-start=\"1776\" data-end=\"1898\">\n<li data-start=\"1776\" data-end=\"1898\">\n<p data-start=\"1778\" data-end=\"1898\">In SSA cases, if [latex]\\sin^{-1}[\/latex] gives an acute angle, check if its supplement also creates a valid triangle.<\/p>\n<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<div>\n<section class=\"textbox example\">\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">In triangle ABC, [latex]A = 35\u00b0[\/latex], [latex]C = 65\u00b0[\/latex], and [latex]c = 10[\/latex] cm. Find side [latex]a[\/latex] and angle [latex]B[\/latex].<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q948825\">Show Solution<\/button><\/p>\n<div id=\"q948825\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">First, find angle [latex]B[\/latex]:<\/p>\n<p>[latex]\\begin{aligned} B &= 180\u00b0 - A - C \\\\ B &= 180\u00b0 - 35\u00b0 - 65\u00b0 \\\\ B &= 80\u00b0 \\end{aligned}[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">Now use the Law of Sines to find side [latex]a[\/latex]:<\/p>\n<p>[latex]\\begin{aligned} \\frac{a}{\\sin A} &= \\frac{c}{\\sin C} \\\\ \\frac{a}{\\sin 35\u00b0} &= \\frac{10}{\\sin 65\u00b0} \\\\ a &= \\frac{10 \\sin 35\u00b0}{\\sin 65\u00b0} \\\\ a &= \\frac{10(0.5736)}{0.9063} \\\\ a &\\approx 6.33 \\text{ cm} \\end{aligned}[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\">Therefore, [latex]B = 80\u00b0[\/latex] and [latex]a \\approx 6.33[\/latex] cm.<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\"><\/div>\n<\/div>\n<\/section>\n<\/div>\n<div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-gfdeddhc-RCyjglaJo5w\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/RCyjglaJo5w?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-gfdeddhc-RCyjglaJo5w\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14661402&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-gfdeddhc-RCyjglaJo5w&#38;vembed=0&#38;video_id=RCyjglaJo5w&#38;video_target=tpm-plugin-gfdeddhc-RCyjglaJo5w\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/The+Ambiguous+Case+for+Sine+Law+-+Nerdstudy_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cThe Ambiguous Case for Sine Law &#8211; Nerdstudy\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<\/div>\n<h2>Area of an Oblique Triangle with Sine<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p data-start=\"66\" data-end=\"360\">When a triangle isn\u2019t a right triangle, we can still find its area using the sine function. Instead of needing the height, we use two sides and the sine of the included angle. This is especially useful for <strong data-start=\"272\" data-end=\"279\">SAS<\/strong> (side\u2013angle\u2013side) cases, where two sides and the angle between them are known.<\/p>\n<p data-start=\"362\" data-end=\"379\">The formula is:<\/p>\n<p data-start=\"381\" data-end=\"478\">[latex]\\text{Area} = \\dfrac{1}{2}ab\\sin C = \\dfrac{1}{2}bc\\sin A = \\dfrac{1}{2}ca\\sin B[\/latex]<\/p>\n<\/div>\n<div>\n<section class=\"textbox example\">\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Find the area of triangle ABC where [latex]a = 8[\/latex] m, [latex]c = 14[\/latex] m, and [latex]B = 55\u00b0[\/latex].<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q566485\">Show Solution<\/button><\/p>\n<div id=\"q566485\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">We have two sides and the included angle between them, so use:<\/p>\n<p>[latex]\\text{Area} = \\frac{1}{2}ac\\sin B[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">Substitute the values:<\/p>\n<p>[latex]\\begin{aligned} \\text{Area} &= \\frac{1}{2}(8)(14)\\sin 55\u00b0 \\\\ &= \\frac{1}{2}(112)\\sin 55\u00b0 \\\\ &= 56 \\sin 55\u00b0 \\\\ &= 56(0.8192) \\\\ &\\approx 45.9 \\text{ m}^2 \\end{aligned}[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\">The area of the triangle is approximately [latex]45.9[\/latex] square meters.<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\"><\/div>\n<\/div>\n<\/section>\n<\/div>\n<div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-chefaaha-mBFDq4bPXMs\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/mBFDq4bPXMs?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-chefaaha-mBFDq4bPXMs\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14661403&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-chefaaha-mBFDq4bPXMs&#38;vembed=0&#38;video_id=mBFDq4bPXMs&#38;video_target=tpm-plugin-chefaaha-mBFDq4bPXMs\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/The+Area+of+a+Triangle+using+Sine_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cThe Area of a Triangle using Sine\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<\/div>\n<h2>Applied Problems with the Law of Sines<\/h2>\n<div>\n<section class=\"textbox example\">\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">Two ranger stations are 15 miles apart, with station A directly east of station B. A fire is spotted from both stations. From station A, the angle to the fire is [latex]42\u00b0[\/latex] west of north. From station B, the angle to the fire is [latex]64\u00b0[\/latex] east of north. How far is the fire from station A?<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-[1.7]\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q76468\">Show Solution<\/button><\/p>\n<div id=\"q76468\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\">Draw a diagram with:<\/p>\n<ul class=\"&#091;li_&amp;&#093;:mb-0 &#091;li_&amp;&#093;:mt-1 &#091;li_&amp;&#093;:gap-1 &#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc flex flex-col gap-1 pl-8 mb-3\">\n<li class=\"whitespace-normal break-words pl-2\">Stations A and B are 15 miles apart<\/li>\n<li class=\"whitespace-normal break-words pl-2\">The fire forms a triangle with the two stations<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-5630\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM.png\" alt=\"The image shows a triangle formed by three points labeled A, B, and Fire. Points A and B lie horizontally along the bottom and are 15 miles apart.\" width=\"344\" height=\"216\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM.png 716w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM-300x189.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM-65x41.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM-225x141.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/07\/13171150\/Screenshot-2026-02-13-at-10.11.19%E2%80%AFAM-350x220.png 350w\" sizes=\"(max-width: 344px) 100vw, 344px\" \/><\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">The angle at station A (interior angle of triangle) is: [latex]90\u00b0 - 42\u00b0 = 48\u00b0[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">The angle at station B (interior angle of triangle) is: [latex]90\u00b0 - 64\u00b0 = 26\u00b0[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">The angle at the fire location is: [latex]180\u00b0 - 48\u00b0 - 26\u00b0 = 106\u00b0[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\">Let [latex]d[\/latex] be the distance from station A to the fire. The side opposite station A (from B to fire) is what we need, and the side opposite the fire (from A to B) is 15 miles.<\/p>\n<p class=\"font-claude-response-body break-words whitespace-pre-wrap leading-&#091;1.7&#093;\">Using the Law of Sines:<\/p>\n<p>[latex]\\begin{aligned} \\frac{15}{\\sin 106\u00b0} &= \\frac{d}{\\sin 26\u00b0} \\\\ d &= \\frac{15 \\sin 26\u00b0}{\\sin 106\u00b0} \\\\ d &= \\frac{15(0.4384)}{0.9613} \\\\ d &\\approx 6.8 \\text{ miles} \\end{aligned}[\/latex]<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\">The fire is approximately [latex]6.8[\/latex] miles from station A.<\/p>\n<p class=\"font-claude-response-body break-words whitespace-normal leading-&#091;1.7&#093;\"><\/div>\n<\/div>\n<\/section>\n<\/div>\n<div>\n<section class=\"textbox watchIt\"><script type=\"text\/javascript\" src=\"https:\/\/www.youtube.com\/iframe_api\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-gcfbabcb-6dEvIdTJ708\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/6dEvIdTJ708?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-gcfbabcb-6dEvIdTJ708\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script type=\"text\/javascript\" src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&#38;cc_minimizable=1&#38;cc_minimize_on_load=0&#38;cc_multi_text_track=0&#38;cc_overlay=1&#38;cc_searchable=0&#38;embed=ajax&#38;mf=14661404&#38;p3sdk_version=1.11.7&#38;p=20361&#38;player_type=youtube&#38;plugin_skin=dark&#38;target=3p-plugin-target-gcfbabcb-6dEvIdTJ708&#38;vembed=0&#38;video_id=6dEvIdTJ708&#38;video_target=tpm-plugin-gcfbabcb-6dEvIdTJ708\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Precalculus\/Transcripts\/Example+-+Application+Problem+Solved+Using+the+Law+of+Sines_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExample: Application Problem Solved Using the Law of Sines\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"author":67,"menu_order":20,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"The Ambiguous Case for Sine Law - 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