{"id":4551,"date":"2025-10-08T15:58:32","date_gmt":"2025-10-08T15:58:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=4551"},"modified":"2025-10-08T20:52:22","modified_gmt":"2025-10-08T20:52:22","slug":"angles-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/angles-learn-it-6\/","title":{"raw":"Angles: Learn It 6","rendered":"Angles: Learn It 6"},"content":{"raw":"

Finding Complementary and Supplementary Angles<\/h2>\r\n
Complementary angles are two angles whose measures add up to 90\u00b0 (or [latex]\\dfrac{\\pi}{2}[\/latex] radians).Supplementary angles are two angles whose measures add up to 180\u00b0 (or [latex]\\pi[\/latex] radians).\r\n\r\n<\/section>
\r\n

Pro Tip:<\/strong> An angle can only have a complement if it measures less than 90\u00b0. An angle of 90\u00b0 or greater has no complement.<\/p>\r\n\r\n<\/section>

\r\n
    \r\n \t
  1. Find the complement of [latex]37\u00b0[\/latex].<\/li>\r\n \t
  2. Find the supplement of [latex]125\u00b0[\/latex].<\/li>\r\n<\/ol>\r\n

    [reveal-answer q=\"994064\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"994064\"]<\/p>\r\n\r\n

      \r\n \t
    1. [latex]\\begin{aligned} \\text{Complement} &= 90\u00b0 - 37\u00b0 \\ &= 53\u00b0 \\end{aligned}[\/latex]The complement of [latex]37\u00b0[\/latex] is [latex]53\u00b0[\/latex].\r\n\r\nWe can verify: [latex]37\u00b0 + 53\u00b0 = 90\u00b0[\/latex]<\/li>\r\n \t
    2. [latex]\\begin{aligned} \\text{Supplement} &= 180\u00b0 - 125\u00b0 \\ &= 55\u00b0 \\end{aligned}[\/latex]\r\n<\/span>\r\nThe supplement of [latex]125\u00b0[\/latex] is [latex]55\u00b0[\/latex].<\/span><\/span>We can verify: [latex]125\u00b0 + 55\u00b0 = 180\u00b0[\/latex] \u2713[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section>
      [ohm_question hide_question_numbers=1]146495[\/ohm_question]<\/section>\r\n

      Finding Complements and Supplements in Radians<\/h2>\r\n

      When working with radians, complementary angles sum to [latex]\\dfrac{\\pi}{2}[\/latex] radians (which equals 90\u00b0) and supplementary angles sum to [latex]\\dfrac{\\pi}[\/latex] radians (which equals 180\u00b0)<\/p>\r\n\r\n

      \r\n
        \r\n \t
      1. \r\n

        Find the complement of [latex]\\dfrac{\\pi}{6}[\/latex] radians.<\/p>\r\n<\/li>\r\n \t

      2. Find the supplement of [latex]\\dfrac{2\\pi}{3}[\/latex] radians.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"882669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"882669\"]\r\n
          \r\n \t
        1. \r\n

          [latex]\\begin{aligned} \\text{Complement} &= \\dfrac{\\pi}{2} - \\dfrac{\\pi}{6} \\ &= \\dfrac{3\\pi}{6} - \\dfrac{\\pi}{6} \\ &= \\dfrac{2\\pi}{6} \\ &= \\dfrac{\\pi}{3} \\end{aligned}[\/latex]<\/p>\r\n

          The complement of [latex]\\dfrac{\\pi}{6}[\/latex] is [latex]\\dfrac{\\pi}{3}[\/latex].<\/p>\r\n

          We can verify: [latex]\\dfrac{\\pi}{6} + \\dfrac{\\pi}{3} = \\dfrac{\\pi}{6} + \\dfrac{2\\pi}{6} = \\dfrac{3\\pi}{6} = \\dfrac{\\pi}{2}[\/latex]<\/p>\r\n<\/li>\r\n \t

        2. \r\n

          [latex]\\begin{aligned} \\text{Supplement} &= \\pi - \\dfrac{2\\pi}{3} \\ &= \\dfrac{3\\pi}{3} - \\dfrac{2\\pi}{3} \\ &= \\dfrac{\\pi}{3} \\end{aligned}[\/latex]<\/p>\r\n

          The supplement of [latex]\\dfrac{2\\pi}{3}[\/latex] is [latex]\\dfrac{\\pi}{3}[\/latex].<\/p>\r\n

          We can verify: [latex]\\dfrac{2\\pi}{3} + \\dfrac{\\pi}{3} = \\dfrac{3\\pi}{3} = \\pi[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>

          \r\n

          How To: Finding Complements in Radians<\/strong><\/p>\r\n\r\n

            \r\n \t
          1. Use the formula: [latex]\\text{Complement} = \\dfrac{\\pi}{2} - \\text{angle}[\/latex]<\/li>\r\n \t
          2. Find a common denominator to subtract the fractions<\/li>\r\n \t
          3. Simplify the result<\/li>\r\n \t
          4. Verify that the sum equals [latex]\\dfrac{\\pi}{2}[\/latex]<\/li>\r\n<\/ol>\r\n

            How To: Finding Supplements in Radians<\/strong><\/p>\r\n\r\n

              \r\n \t
            1. Use the formula: [latex]\\text{Complement} = \\dfrac{\\pi} - \\text{angle}[\/latex]<\/li>\r\n \t
            2. Find a common denominator to subtract the fractions<\/li>\r\n \t
            3. Simplify the result<\/li>\r\n \t
            4. Verify that the sum equals [latex]\\dfrac{\\pi}[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>
              [ohm_question hide_question_numbers=1]313397[\/ohm_question]<\/section>\r\n

              Finding Complements and Supplements in DMS Form<\/h2>\r\n
              1\u00b0 = 60' (minutes) and 1' = 60\" (seconds).<\/section>
              \r\n
                \r\n \t
              1. Find the complement of [latex]35\u00b0 20' 15\"[\/latex].<\/li>\r\n \t
              2. Find the supplement of [latex]72\u00b0 45' 20\"[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"664386\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"664386\"]\r\n
                  \r\n \t
                1. Find the complement of [latex]35\u00b0 20' 15\"[\/latex].\r\n

                  We need to subtract [latex]35\u00b0 20' 15\"[\/latex] from [latex]90\u00b0 0' 0\"[\/latex].<\/p>\r\n

                  Rewrite [latex]90\u00b0 0' 0\"[\/latex] as [latex]89\u00b0 60' 0\"[\/latex], then as [latex]89\u00b0 59' 60\"[\/latex]:<\/p>\r\n

                  [latex]\\begin{aligned} &\\phantom{-}89\u00b0 59' 60\" \\ &\\underline{-35\u00b0 20' 15\"} \\ &\\phantom{-}54\u00b0 39' 45\" \\end{aligned}[\/latex]<\/p>\r\n

                  Working from right to left:<\/p>\r\n\r\n

                    \r\n \t
                  • Seconds: [latex]60\" - 15\" = 45\"[\/latex]<\/li>\r\n \t
                  • Minutes: [latex]59' - 20' = 39'[\/latex]<\/li>\r\n \t
                  • Degrees: [latex]89\u00b0 - 35\u00b0 = 54\u00b0[\/latex]<\/li>\r\n<\/ul>\r\n

                    The complement of [latex]35\u00b0 20' 15\"[\/latex] is [latex]54\u00b0 39' 45\"[\/latex].<\/p>\r\n<\/li>\r\n \t

                  • Find the supplement of [latex]72\u00b0 45' 20\"[\/latex].\r\n

                    We need to subtract [latex]72\u00b0 45' 20\"[\/latex] from [latex]180\u00b0 0' 0\"[\/latex].<\/p>\r\n

                    Rewrite [latex]180\u00b0 0' 0\"[\/latex] as [latex]179\u00b0 60' 0\"[\/latex], then as [latex]179\u00b0 59' 60\"[\/latex]:<\/p>\r\n

                    [latex]\\begin{aligned} &\\phantom{-}179\u00b0 59' 60\" \\ &\\underline{-\\phantom{1}72\u00b0 45' 20\"} \\ &\\phantom{-}107\u00b0 14' 40\" \\end{aligned}[\/latex]<\/p>\r\n

                    Working from right to left:<\/p>\r\n\r\n

                      \r\n \t
                    • Seconds: [latex]60\" - 20\" = 40\"[\/latex]<\/li>\r\n \t
                    • Minutes: [latex]59' - 45' = 14'[\/latex]<\/li>\r\n \t
                    • Degrees: [latex]179\u00b0 - 72\u00b0 = 107\u00b0[\/latex]<\/li>\r\n<\/ul>\r\n

                      The supplement of [latex]72\u00b0 45' 20\"[\/latex] is [latex]107\u00b0 14' 40\"[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>

                      \r\n

                      How To: Finding Complements or Supplements in DMS Form<\/strong><\/p>\r\n\r\n

                        \r\n \t
                      1. Write 90\u00b0 in DMS form as [latex]89\u00b0 59' 60\"[\/latex] (this makes subtraction easier) or write 180\u00b0 as [latex]159\u00b0 59' 60\"[\/latex]<\/li>\r\n \t
                      2. Subtract seconds from seconds, minutes from minutes, degrees from degrees<\/li>\r\n \t
                      3. Simplify your answer<\/li>\r\n<\/ol>\r\n<\/section>
                        If you need to borrow, remember: 1\u00b0 = 60' and 1' = 60\"\r\n\r\n<\/section>
                        [ohm_question hide_question_numbers=1]313398[\/ohm_question]<\/section>","rendered":"

                        Finding Complementary and Supplementary Angles<\/h2>\n
                        Complementary angles are two angles whose measures add up to 90\u00b0 (or [latex]\\dfrac{\\pi}{2}[\/latex] radians).Supplementary angles are two angles whose measures add up to 180\u00b0 (or [latex]\\pi[\/latex] radians).<\/p>\n<\/section>\n
                        \n

                        Pro Tip:<\/strong> An angle can only have a complement if it measures less than 90\u00b0. An angle of 90\u00b0 or greater has no complement.<\/p>\n<\/section>\n

                        \n
                          \n
                        1. Find the complement of [latex]37\u00b0[\/latex].<\/li>\n
                        2. Find the supplement of [latex]125\u00b0[\/latex].<\/li>\n<\/ol>\n

                          \n