{"id":2211,"date":"2025-08-11T16:53:30","date_gmt":"2025-08-11T16:53:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2211"},"modified":"2025-08-13T16:56:18","modified_gmt":"2025-08-13T16:56:18","slug":"non-right-triangles-with-law-of-cosines-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-cosines-learn-it-2\/","title":{"raw":"Non-right Triangles with Law of Cosines: Learn It 2","rendered":"Non-right Triangles with Law of Cosines: Learn It 2"},"content":{"raw":"

Solving Triangles with the Law of Cosines<\/h2>\r\n
How To:\u00a0Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/strong>\r\n
    \r\n \t
  1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.<\/li>\r\n \t
  2. Apply the Law of Cosines to find the length of the unknown side or angle.<\/li>\r\n \t
  3. Apply the Law of Sines<\/strong> or Cosines to find the measure of a second angle.<\/li>\r\n \t
  4. Compute the measure of the remaining angle.<\/li>\r\n<\/ol>\r\n<\/section>
    Find the unknown side and angles of the triangle.\"A\r\n\r\n[reveal-answer q=\"359338\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359338\"]\r\n\r\nFirst, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.\r\n\r\nEach one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side [latex]b[\/latex], as we know the measurement of the opposite angle [latex]\\beta [\/latex].\r\n

    [latex]\\begin{align}&{b}^{2}={a}^{2}+{c}^{2}-2ac\\cos \\beta \\\\ &{b}^{2}={10}^{2}+{12}^{2}-2\\left(10\\right)\\left(12\\right)\\cos \\left({30}^{\\circ }\\right) && \\text{Substitute the measurements for the known quantities}. \\\\ &{b}^{2}=100+144 - 240\\left(\\frac{\\sqrt{3}}{2}\\right)&& \\text{Evaluate the cosine and begin to simplify}. \\\\ &{b}^{2}=244 - 120\\sqrt{3} \\\\ &b=\\sqrt{244 - 120\\sqrt{3}}&& \\text{Use the square root property}. \\\\ &b\\approx 6.013 \\end{align}[\/latex]<\/p>\r\nBecause we are solving for a length, we use only the positive square root. Now that we know the length [latex]b[\/latex], we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle [latex]\\alpha [\/latex], we have\r\n

    [latex]\\begin{align}&\\frac{\\sin \\alpha }{a}=\\frac{\\sin \\beta }{b} \\\\ &\\frac{\\sin \\alpha }{10}=\\frac{\\sin \\left(30^\\circ \\right)}{6.013} \\\\ &\\sin \\alpha =\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}&& \\text{Multiply both sides of the equation by 10}. \\\\ &\\alpha ={\\sin }^{-1}\\left(\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}\\right)&& \\text{Find the inverse sine of }\\frac{10\\sin \\left(30^\\circ \\right)}{6.013}. \\\\ &\\alpha \\approx 56.3^\\circ \\end{align}[\/latex]<\/p>\r\nThe other possibility for [latex]\\alpha [\/latex] would be [latex]\\alpha =180^\\circ -56.3^\\circ \\approx 123.7^\\circ [\/latex]. In the original diagram, [latex]\\alpha [\/latex] is adjacent to the longest side, so [latex]\\alpha [\/latex] is an acute angle and, therefore, [latex]123.7^\\circ [\/latex] does not make sense. Notice that if we choose to apply the Law of Cosines<\/strong>, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between [latex]0^\\circ [\/latex] and [latex]180^\\circ [\/latex]. Proceeding with [latex]\\alpha \\approx 56.3^\\circ [\/latex], we can then find the third angle of the triangle.\r\n

    [latex]\\gamma =180^\\circ -30^\\circ -56.3^\\circ \\approx 93.7^\\circ [\/latex]<\/p>\r\nThe complete set of angles and sides is\r\n

    [latex]\\begin{align}&\\alpha \\approx 56.3^\\circ && a=10\\\\ &\\beta =30^\\circ && b\\approx 6.013\\\\ &\\gamma \\approx 93.7^\\circ && c=12 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    \r\n
    \r\n\r\nFind the missing side and angles of the given triangle: [latex]\\alpha =30^\\circ ,b=12,c=24[\/latex].\r\n\r\n[reveal-answer q=\"885435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"885435\"]\r\n\r\n[latex]a\\approx 14.9,\\beta \\approx 23.8^\\circ ,\\gamma \\approx 126.2^\\circ [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>
    [ohm_question hide_question_numbers=1]174837[\/ohm_question]<\/section>
    Find the angle [latex]\\alpha [\/latex] for the given triangle if side [latex]a=20[\/latex], side [latex]b=25[\/latex], and side [latex]c=18[\/latex].[reveal-answer q=\"211238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"211238\"]\r\n\r\nFor this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle [latex]\\alpha [\/latex], we have\r\n

    [latex]\\begin{align} &{a}^{2}={b}^{2}+{c}^{2}-2bc\\cos \\alpha \\\\ &{20}^{2}={25}^{2}+{18}^{2}-2\\left(25\\right)\\left(18\\right)\\cos \\alpha&& \\text{Substitute the appropriate measurements}. \\\\ &400=625+324 - 900\\cos \\alpha&& \\text{Simplify in each step}. \\\\ &400=949 - 900\\cos \\alpha \\\\ &-549=-900\\cos \\alpha&& \\text{Isolate cos }\\alpha . \\\\ &\\frac{-549}{-900}=\\cos \\alpha \\\\ &0.61\\approx \\cos \\alpha \\\\ &{\\cos }^{-1}\\left(0.61\\right)\\approx \\alpha&& \\text{Find the inverse cosine}. \\\\ &\\alpha \\approx 52.4^\\circ \\end{align}[\/latex]<\/p>\r\n \r\n\r\n\"A\r\n

    Analysis of the Solution<\/h4>\r\nBecause the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
    Given [latex]a=5,b=7[\/latex], and [latex]c=10[\/latex], find the missing angles.[reveal-answer q=\"488277\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"488277\"][latex]\\alpha \\approx 27.7^\\circ ,\\beta \\approx 40.5^\\circ ,\\gamma \\approx 111.8^\\circ [\/latex][\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]173789[\/ohm_question]<\/section>","rendered":"

    Solving Triangles with the Law of Cosines<\/h2>\n
    How To:\u00a0Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.<\/strong><\/p>\n
      \n
    1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.<\/li>\n
    2. Apply the Law of Cosines to find the length of the unknown side or angle.<\/li>\n
    3. Apply the Law of Sines<\/strong> or Cosines to find the measure of a second angle.<\/li>\n
    4. Compute the measure of the remaining angle.<\/li>\n<\/ol>\n<\/section>\n
      Find the unknown side and angles of the triangle.\"A<\/p>\n