{"id":2123,"date":"2025-08-04T18:14:13","date_gmt":"2025-08-04T18:14:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2123"},"modified":"2025-12-02T22:43:54","modified_gmt":"2025-12-02T22:43:54","slug":"non-right-triangles-with-law-of-sines-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-sines-learn-it-2\/","title":{"raw":"Non-right Triangles with Law of Sines: Learn It 2","rendered":"Non-right Triangles with Law of Sines: Learn It 2"},"content":{"raw":"

Using The Law of Sines to Solve SSA Triangles<\/h2>\r\nWe can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case<\/strong>. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.\r\n\r\n
\r\n

the ambiguous case<\/h3>\r\nOblique triangles in the category SSA may have four different outcomes.\u00a0Figure 9\u00a0illustrates the solutions with the known sides [latex]a[\/latex] and [latex]b[\/latex] and known angle [latex]\\alpha [\/latex].\r\n\r\n\"Four\r\n\r\n<\/section>
Solve the triangle for the missing side and find the missing angle measures to the nearest tenth.\"An[reveal-answer q=\"588123\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"588123\"]Use the Law of Sines to find angle [latex]\\beta [\/latex] and angle [latex]\\gamma [\/latex], and then side [latex]c[\/latex]. Solving for [latex]\\beta [\/latex], we have the proportion\r\n

[latex]\\begin{gathered} \\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ \\frac{\\sin \\left(35^\\circ \\right)}{6}=\\frac{\\sin \\beta }{8}\\\\ \\frac{8\\sin \\left(35^\\circ \\right)}{6}=\\sin \\beta \\\\ 0.7648\\approx \\sin \\beta \\\\ {\\sin }^{-1}\\left(0.7648\\right)\\approx 49.9^\\circ \\\\ \\beta \\approx 49.9^\\circ \\end{gathered}[\/latex]<\/p>\r\nHowever, in the diagram, angle [latex]\\beta [\/latex] appears to be an obtuse angle and may be greater than 90\u00b0. How did we get an acute angle, and how do we find the measurement of [latex]\\beta ?[\/latex] Let\u2019s investigate further. Dropping a perpendicular from [latex]\\gamma [\/latex] and viewing the triangle from a right angle perspective, it appears that there may be a second triangle that will fit the given criteria.\r\n\r\n\"An\r\n\r\nThe angle supplementary to [latex]\\beta [\/latex] is approximately equal to 49.9\u00b0, which means that [latex]\\beta =180^\\circ -49.9^\\circ =130.1^\\circ [\/latex]. (Remember that the sine function is positive in both the first and second quadrants.) Solving for [latex]\\gamma [\/latex], we have\r\n

[latex]\\gamma =180^\\circ -35^\\circ -130.1^\\circ \\approx 14.9^\\circ [\/latex]<\/p>\r\nWe can then use these measurements to solve the other triangle. Since [latex]{\\gamma }^{\\prime }[\/latex] is supplementary to [latex]\\gamma [\/latex], we have\r\n

[latex]{\\gamma }^{\\prime }=180^\\circ -35^\\circ -49.9^\\circ \\approx 95.1^\\circ [\/latex]<\/p>\r\nNow we need to find [latex]c[\/latex] and [latex]{c}^{\\prime }[\/latex].\r\n\r\nWe have\r\n

[latex]\\begin{gathered}\\frac{c}{\\sin \\left(14.9^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ c=\\frac{6\\sin \\left(14.9^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 2.7 \\end{gathered}[\/latex]<\/p>\r\nFinally,\r\n

[latex]\\begin{gathered}\\frac{{c}^{\\prime }}{\\sin \\left(95.1^\\circ \\right)}=\\frac{6}{\\sin \\left(35^\\circ \\right)} \\\\ {c}^{\\prime }=\\frac{6\\sin \\left(95.1^\\circ \\right)}{\\sin \\left(35^\\circ \\right)}\\approx 10.4 \\end{gathered}[\/latex]<\/p>\r\nTo summarize, there are two triangles with an angle of 35\u00b0, an adjacent side of 8, and an opposite side of 6.\r\n\r\n\"There\r\n\r\nHowever, we were looking for the values for the triangle with an obtuse angle [latex]\\beta [\/latex]. We can see them in the first triangle (a) in Figure 12.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Given [latex]\\alpha =80^\\circ ,a=120[\/latex], and [latex]b=121[\/latex], find the missing side and angles. If there is more than one possible solution, show both.[reveal-answer q=\"65799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65799\"]Solution 1<\/strong>\r\n[latex]\\begin{align}&\\alpha =80^\\circ && a=120\\hfill \\\\ &\\beta \\approx 83.2^\\circ && b=121 \\\\ &\\gamma \\approx 16.8^\\circ && c\\approx 35.2 \\end{align}[\/latex]\r\nSolution 2<\/strong>\r\n[latex]\\begin{align}&{\\alpha }^{\\prime }=80^\\circ &&{a}^{\\prime }=120 \\\\ &{\\beta }^{\\prime }\\approx 96.8^\\circ &&{b}^{\\prime }=121 \\\\ &{\\gamma }^{\\prime }\\approx 3.2^\\circ &&{c}^{\\prime }\\approx 6.8 \\end{align}[\/latex][\/hidden-answer]<\/section>
Solve for the unknown side and angles. Round your answers to the nearest tenth.\"An[reveal-answer q=\"357327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"357327\"]In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle [latex]\\gamma =85^\\circ [\/latex], and its corresponding side [latex]c=12[\/latex], and we know side [latex]b=9[\/latex]. We will use this proportion to solve for [latex]\\beta [\/latex].\r\n

[latex]\\begin{align}\\frac{\\sin \\left(85^\\circ \\right)}{12}&=\\frac{\\sin \\beta }{9} && \\text{Isolate the unknown}.\\\\ \\frac{9\\sin \\left(85^\\circ \\right)}{12}&=\\sin \\beta \\end{align}[\/latex]<\/p>\r\nTo find [latex]\\beta [\/latex], apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for [latex]\\beta [\/latex]. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.\r\n

[latex]\\begin{align}\\beta &={\\sin }^{-1}\\left(\\frac{9\\sin \\left(85^\\circ \\right)}{12}\\right) \\\\ \\beta &\\approx {\\sin }^{-1}\\left(0.7471\\right) \\\\ \\beta &\\approx 48.3^\\circ \\end{align}[\/latex]<\/p>\r\nIn this case, if we subtract [latex]\\beta [\/latex] from 180\u00b0, we find that there may be a second possible solution. Thus, [latex]\\beta =180^\\circ -48.3^\\circ \\approx 131.7^\\circ [\/latex]. To check the solution, subtract both angles, 131.7\u00b0 and 85\u00b0, from 180\u00b0. This gives\r\n

[latex]\\alpha =180^\\circ -85^\\circ -131.7^\\circ \\approx -36.7^\\circ [\/latex],<\/p>\r\nwhich is impossible, and so [latex]\\beta \\approx 48.3^\\circ [\/latex].\r\n\r\nTo find the remaining missing values, we calculate [latex]\\alpha =180^\\circ -85^\\circ -48.3^\\circ \\approx 46.7^\\circ [\/latex]. Now, only side [latex]a[\/latex] is needed. Use the Law of Sines to solve for [latex]a[\/latex] by one of the proportions.\r\n

[latex]\\begin{gathered} \\frac{\\sin \\left(85^\\circ \\right)}{12}=\\frac{\\sin \\left(46.7^\\circ \\right)}{a} \\\\ a\\frac{\\sin \\left(85^\\circ \\right)}{12}=\\sin \\left(46.7^\\circ \\right) \\\\ a=\\frac{12\\sin \\left(46.7^\\circ \\right)}{\\sin \\left(85^\\circ \\right)}\\approx 8.8 \\end{gathered}[\/latex]<\/p>\r\nThe complete set of solutions for the given triangle is\r\n

[latex]\\begin{gathered} \\alpha \\approx 46.7^\\circ \\text{, }a\\approx 8.8 \\\\ \\beta \\approx 48.3^\\circ \\text{, }b=9 \\\\ \\gamma =85^\\circ \\text{, }c=12\\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Given [latex]\\alpha =80^\\circ ,a=100,b=10[\/latex], find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.[reveal-answer q=\"396947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"396947\"][latex]\\beta \\approx 5.7^\\circ ,\\gamma \\approx 94.3^\\circ ,c\\approx 101.3[\/latex][\/hidden-answer]<\/section>
Find all possible triangles if one side has length 4 opposite an angle of 50\u00b0, and a second side has length 10.[reveal-answer q=\"979773\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979773\"]Using the given information, we can solve for the angle opposite the side of length 10.\r\n

[latex]\\begin{gathered}\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha =\\frac{10\\sin \\left(50^\\circ \\right)}{4} \\\\ \\sin \\alpha \\approx 1.915 \\end{gathered}[\/latex]<\/p>\r\n\"An\r\n\r\nWe can stop here without finding the value of [latex]\\alpha [\/latex]. Because the range of the sine function is [latex]\\left[-1,1\\right][\/latex], it is impossible for the sine value to be 1.915. In fact, inputting [latex]{\\sin }^{-1}\\left(1.915\\right)[\/latex] in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Determine the number of triangles possible given [latex]a=31,b=26,\\beta =48^\\circ [\/latex].[reveal-answer q=\"996706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"996706\"]Two[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]149233[\/ohm_question]<\/section>","rendered":"

Using The Law of Sines to Solve SSA Triangles<\/h2>\n

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case<\/strong>. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.<\/p>\n

\n

the ambiguous case<\/h3>\n

Oblique triangles in the category SSA may have four different outcomes.\u00a0Figure 9\u00a0illustrates the solutions with the known sides [latex]a[\/latex] and [latex]b[\/latex] and known angle [latex]\\alpha[\/latex].<\/p>\n

\"Four<\/p>\n<\/section>\n

Solve the triangle for the missing side and find the missing angle measures to the nearest tenth.\"An<\/p>\n