{"id":1562,"date":"2021-12-02T19:38:30","date_gmt":"2021-12-03T00:38:30","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/11-8-sine-and-cosine-laws\/"},"modified":"2023-08-31T19:29:34","modified_gmt":"2023-08-31T23:29:34","slug":"sine-and-cosine-laws","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/sine-and-cosine-laws\/","title":{"raw":"11.8 Sine and Cosine Laws","rendered":"11.8 Sine and Cosine Laws"},"content":{"raw":"Right angle trigonometry is generally limited to triangles that contain a right angle. It is possible to use trigonometry with non-right triangles using two laws: the sine law and the cosine law.\r\n

The Law of Sines<\/h1>\r\nThe sine law is a ratio of sines and opposite sides. The law takes the following form:\r\n\r\n[latex]\\dfrac{a}{\\text{sin }A}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{b}{\\text{sin }B}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{c}{\\text{sin }C}[\/latex]\r\n\r\nSometimes, it is written and used as the reciprocal of the above:\r\n\r\n[latex]\\dfrac{\\text{sin }A}{a}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{\\text{sin }B}{b}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{\\text{sin }C}{c}[\/latex]\r\n\r\n\"The\r\n\r\nThe law of sine is used when either two sides and one opposite angle of one of the sides are known, or when there are two angles and one side of one of the angles. If there are two given angles of a triangle, then all three angles are known, since [latex]A^{\\circ} + B^{\\circ} + C^{\\circ} = 180^{\\circ}.[\/latex]\r\n\r\nThe sine law is a very useful law with one caveat in that it is possible to sometimes have two triangles (one larger and one smaller) that generate the same result. This is termed the ambiguous case and is described later in this section.\r\n\r\nThere are also textbook errors where the data given for the triangle is impossible to create. For instance:\r\n
\r\n

Example 11.8.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nCan the following triangle exist?\r\n\r\n\"Triangle\r\n\r\nIf this triangle can exist, then the ratio of sines for the angles to the opposite sides should equate.\r\n\r\n[latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in} = \\hspace{0.25in} \\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in} = \\hspace{0.25in} \\dfrac{10}{\\text{sin }120^{\\circ}}[\/latex]\r\n\r\nReducing this yields:\r\n\r\n[latex]\\dfrac{6}{0.5} \\hspace{0.25in} = \\hspace{0.25in}\\dfrac{6}{0.5} \\hspace{0.25in} = \\hspace{0.25in} \\dfrac{10}{0.866}[\/latex]\r\n\r\nIn checking this out, we find that 12 = 12 \u2260 11.55.\r\n\r\nThis means that this triangle cannot exist.\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 11.8.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the correct length of the side opposite 120\u00b0 in the triangle shown below.\r\n\r\n\"Triangle\r\n\r\nFor this triangle, the ratio to solve is:\r\n\r\n[latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{6}{\\text{sin }30^{\\circ}} \\hspace{0.25in}=\\hspace{0.25in}\\dfrac{x}{\\text{sin }120^{\\circ}} [\/latex]\r\n\r\nWe only need to use one portion of this, so:\r\n\r\n[latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{x}{\\text{sin }120^{\\circ}}[\/latex]\r\n\r\nMultiplying both sides of this by sin 120\u00b0, we are left with:\r\n\r\n[latex]x=\\dfrac{6\\text{ sin }120^{\\circ}}{\\text{sin }30^{\\circ}}[\/latex]\r\n\r\nThis leaves us with [latex]x = 10.29[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 11.8.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the correct length of the side opposite 120\u00b0 in the triangle shown below.\r\n\r\n\"Traingle\r\n\r\nFor this triangle, the ratio to solve is:\r\n\r\n[latex]\\dfrac{a}{\\text{sin }44^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{110}{\\text{sin }86^{\\circ}}[\/latex]\r\n\r\nMultiplying both sides by sin 44\u00b0 leaves us with:\r\n\r\n[latex]a=\\dfrac{110\\text{ sin }44^{\\circ}}{\\text{sin }86^{\\circ}}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 11.8.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the unknown angle shown in the triangle shown below.\r\n\r\n\"Traingle\r\n\r\nFor this triangle, the ratio to solve is:\r\n\r\n[latex]\\dfrac{14}{\\text{sin }A}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{16}{\\text{sin }52^{\\circ}}[\/latex]\r\n\r\nIsolating sin A yields:\r\n\r\n[latex]\\text{sin }A=\\dfrac{14\\text{ sin }52^{\\circ}}{16}[\/latex]\r\n\r\nWe now need to take the inverse sin of both sides to solve for A:\r\n\r\n[latex]\\begin{array}{l}\r\nA=\\text{sin}^{-1}\\left(\\dfrac{14\\text{ sin }52^{\\circ}}{16}\\right) \\\\ \\\\\r\nA=43.6^{\\circ}\r\n\\end{array}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n

The Ambiguous Case<\/h2>\r\nIt is possible, when given the right data, to create two different triangles.\r\n\r\n\"1\r\n\r\nYou can see from the triangle shown above that it is possible to have two angles, B1<\/sub> and B2<\/sub>, for side [latex]b[\/latex]. Using the sine law, you will always end up solving for B1<\/sub>, the angle for the largest triangle. If you are trying to solve for the smaller triangle, then you only need to subtract B1<\/sub> from 180\u00b0.\r\n\r\nFor example, if B1<\/sub> = 50\u00b0, then B2<\/sub> = 180\u00b0 \u2212 B1<\/sub>. This means B2<\/sub> = 180\u00b0 \u2212 50\u00b0 or 130\u00b0.\r\n\r\nIf the angle you solve for when using the sine law is smaller than it should be, then correct for it as we just did above.\r\n

The Law of Cosines<\/h1>\r\nThe Law of Cosines is the generalized law of the Pythagorean Theorem [latex](a^2 + b^2 = c^2).[\/latex]\r\n\r\nThe Law of Cosines is generally written in three different forms, which are as follows:\r\n

[latex]\\begin{array}{l}\r\na^2 = b^2 + c^2 - 2bc\\text{ cos }A \\\\\r\nb^2 = a^2 + c^2 - 2ac\\text{ cos }B \\\\\r\nc^2 = a^2 + b^2 - 2ab\\text{ cos }C\r\n\\end{array}[\/latex]<\/p>\r\nAll forms revert back to one of the three regular forms of the Pythagorean theorem [latex](a^2 = b^2 + c^2, b^2 = a^2 + c^2, c^2 = a^2 + b^2)[\/latex] if [latex]A, B[\/latex] or [latex]C[\/latex] is 90\u00b0, since [latex]\\text{cos }90^{\\circ} = 0.[\/latex] The following examples illustrate the usage of the cosine law in trigonometry\r\n

\r\n

Example 11.8.5<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the unknown angle shown in the triangle shown below.\r\n\r\n\"Triangle\r\n\r\nFor this triangle:\r\n

[latex]\\begin{array}{l}\r\nA=\\text{find} \\\\ \\\\\r\na=90 \\\\ \\\\\r\nb=50 \\\\ \\\\\r\nc=100\r\n\\end{array}[\/latex]<\/p>\r\nNote that [latex]b[\/latex] and [latex]c[\/latex] could be switched around. Now, to solve:\r\n

[latex]\\begin{array}{rrlllll}\r\na^2 &= &\\phantom{-}b^2 &+ &c^2 &-& 2bc\\text{ cos }A \\\\ \\\\\r\n90^2 &= &\\phantom{-}50^2 &+ &100^2 &- &2(50)(100)\\text{ cos } A \\\\\r\n8100& =&\\phantom{-}2500 &+ &10000& - &10000\\text{ cos }A \\\\\r\n-2500&&-2500&-&10000&& \\\\\r\n\\hline\r\n-4400& =&-10000&\\text{ cos }&A&& \\\\ \\\\\r\n\\text{ cos }A& =&\\dfrac{-4400}{-10000}&&&& \\\\ \\\\\r\nB& =&\\text{cos}^{-1}0.44&&&& \\\\ \\\\\r\nB&=&63.9^{\\circ}&&&&\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 11.8.6<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the unknown side shown in the triangle shown below.\r\n\r\n\"TriangleFor this triangle:\r\n

[latex]\\begin{array}{l}\r\nB=70^{\\circ} \\\\ \\\\\r\na=95 \\\\ \\\\\r\nb=\\text{find} \\\\ \\\\\r\nc=60\r\n\\end{array}[\/latex]<\/p>\r\nNote that [latex]b[\/latex] and [latex]c[\/latex] could be switched around. Now, to solve:\r\n

[latex]\\begin{array}{l}\r\nb^2=a^2+c^2-2ac\\text{ cos }B \\\\ \\\\\r\nb^2=95^2+60^2-2(95)(60)\\text{ cos }70^{\\circ} \\\\\r\nb^2=9025+3600-11400(0.34202) \\\\\r\nb^2=12625 - 3899 \\\\\r\nb^2=8726 \\\\ \\\\\r\nb=\\sqrt{8726} \\\\\r\nb=93.4\r\n\\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nUnlike with the law of sines, there should be no ambiguous cases with the law of cosines.\r\n

Questions<\/h1>\r\nSolve all unknowns in the following non-right triangles using either the law of sines or cosines.\r\n
    \r\n \t
  1. \"Traingle<\/li>\r\n \t
  2. \"Triangle<\/li>\r\n \t
  3. \"Triangle<\/li>\r\n \t
  4. \"Triangle<\/li>\r\n \t
  5. \"Traingle<\/li>\r\n \t
  6. \"Traingle<\/li>\r\n \t
  7. \"Triangle<\/li>\r\n \t
  8. \"Triangle<\/li>\r\n \t
  9. \"Traingle<\/li>\r\n \t
  10. \"Triangle<\/li>\r\n \t
  11. \"Triangle<\/li>\r\n \t
  12. \"Triangle<\/li>\r\n \t
  13. \"Triangle<\/li>\r\n \t
  14. \"Triangle<\/li>\r\n \t
  15. \"Triangle<\/li>\r\n \t
  16. \"Triangle<\/li>\r\n<\/ol>\r\nAnswer Key 11.8<\/a>","rendered":"

    Right angle trigonometry is generally limited to triangles that contain a right angle. It is possible to use trigonometry with non-right triangles using two laws: the sine law and the cosine law.<\/p>\n

    The Law of Sines<\/h1>\n

    The sine law is a ratio of sines and opposite sides. The law takes the following form:<\/p>\n

    [latex]\\dfrac{a}{\\text{sin }A}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{b}{\\text{sin }B}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{c}{\\text{sin }C}[\/latex]<\/p>\n

    Sometimes, it is written and used as the reciprocal of the above:<\/p>\n

    [latex]\\dfrac{\\text{sin }A}{a}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{\\text{sin }B}{b}\\hspace{0.25in} =\\hspace{0.25in} \\dfrac{\\text{sin }C}{c}[\/latex]<\/p>\n

    \"The<\/p>\n

    The law of sine is used when either two sides and one opposite angle of one of the sides are known, or when there are two angles and one side of one of the angles. If there are two given angles of a triangle, then all three angles are known, since [latex]A^{\\circ} + B^{\\circ} + C^{\\circ} = 180^{\\circ}.[\/latex]<\/p>\n

    The sine law is a very useful law with one caveat in that it is possible to sometimes have two triangles (one larger and one smaller) that generate the same result. This is termed the ambiguous case and is described later in this section.<\/p>\n

    There are also textbook errors where the data given for the triangle is impossible to create. For instance:<\/p>\n

    \n
    \n

    Example 11.8.1<\/p>\n<\/header>\n

    \n

    Can the following triangle exist?<\/p>\n

    \"Triangle<\/p>\n

    If this triangle can exist, then the ratio of sines for the angles to the opposite sides should equate.<\/p>\n

    [latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in} = \\hspace{0.25in} \\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in} = \\hspace{0.25in} \\dfrac{10}{\\text{sin }120^{\\circ}}[\/latex]<\/p>\n

    Reducing this yields:<\/p>\n

    [latex]\\dfrac{6}{0.5} \\hspace{0.25in} = \\hspace{0.25in}\\dfrac{6}{0.5} \\hspace{0.25in} = \\hspace{0.25in} \\dfrac{10}{0.866}[\/latex]<\/p>\n

    In checking this out, we find that 12 = 12 \u2260 11.55.<\/p>\n

    This means that this triangle cannot exist.<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.8.2<\/p>\n<\/header>\n

    \n

    Find the correct length of the side opposite 120\u00b0 in the triangle shown below.<\/p>\n

    \"Triangle<\/p>\n

    For this triangle, the ratio to solve is:<\/p>\n

    [latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{6}{\\text{sin }30^{\\circ}} \\hspace{0.25in}=\\hspace{0.25in}\\dfrac{x}{\\text{sin }120^{\\circ}}[\/latex]<\/p>\n

    We only need to use one portion of this, so:<\/p>\n

    [latex]\\dfrac{6}{\\text{sin }30^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{x}{\\text{sin }120^{\\circ}}[\/latex]<\/p>\n

    Multiplying both sides of this by sin 120\u00b0, we are left with:<\/p>\n

    [latex]x=\\dfrac{6\\text{ sin }120^{\\circ}}{\\text{sin }30^{\\circ}}[\/latex]<\/p>\n

    This leaves us with [latex]x = 10.29[\/latex].<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.8.3<\/p>\n<\/header>\n

    \n

    Find the correct length of the side opposite 120\u00b0 in the triangle shown below.<\/p>\n

    \"Traingle<\/p>\n

    For this triangle, the ratio to solve is:<\/p>\n

    [latex]\\dfrac{a}{\\text{sin }44^{\\circ}}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{110}{\\text{sin }86^{\\circ}}[\/latex]<\/p>\n

    Multiplying both sides by sin 44\u00b0 leaves us with:<\/p>\n

    [latex]a=\\dfrac{110\\text{ sin }44^{\\circ}}{\\text{sin }86^{\\circ}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.8.4<\/p>\n<\/header>\n

    \n

    Find the unknown angle shown in the triangle shown below.<\/p>\n

    \"Traingle<\/p>\n

    For this triangle, the ratio to solve is:<\/p>\n

    [latex]\\dfrac{14}{\\text{sin }A}\\hspace{0.25in}=\\hspace{0.25in}\\dfrac{16}{\\text{sin }52^{\\circ}}[\/latex]<\/p>\n

    Isolating sin A yields:<\/p>\n

    [latex]\\text{sin }A=\\dfrac{14\\text{ sin }52^{\\circ}}{16}[\/latex]<\/p>\n

    We now need to take the inverse sin of both sides to solve for A:<\/p>\n

    [latex]\\begin{array}{l} A=\\text{sin}^{-1}\\left(\\dfrac{14\\text{ sin }52^{\\circ}}{16}\\right) \\\\ \\\\ A=43.6^{\\circ} \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    The Ambiguous Case<\/h2>\n

    It is possible, when given the right data, to create two different triangles.<\/p>\n

    \"1<\/p>\n

    You can see from the triangle shown above that it is possible to have two angles, B1<\/sub> and B2<\/sub>, for side [latex]b[\/latex]. Using the sine law, you will always end up solving for B1<\/sub>, the angle for the largest triangle. If you are trying to solve for the smaller triangle, then you only need to subtract B1<\/sub> from 180\u00b0.<\/p>\n

    For example, if B1<\/sub> = 50\u00b0, then B2<\/sub> = 180\u00b0 \u2212 B1<\/sub>. This means B2<\/sub> = 180\u00b0 \u2212 50\u00b0 or 130\u00b0.<\/p>\n

    If the angle you solve for when using the sine law is smaller than it should be, then correct for it as we just did above.<\/p>\n

    The Law of Cosines<\/h1>\n

    The Law of Cosines is the generalized law of the Pythagorean Theorem [latex](a^2 + b^2 = c^2).[\/latex]<\/p>\n

    The Law of Cosines is generally written in three different forms, which are as follows:<\/p>\n

    [latex]\\begin{array}{l} a^2 = b^2 + c^2 - 2bc\\text{ cos }A \\\\ b^2 = a^2 + c^2 - 2ac\\text{ cos }B \\\\ c^2 = a^2 + b^2 - 2ab\\text{ cos }C \\end{array}[\/latex]<\/p>\n

    All forms revert back to one of the three regular forms of the Pythagorean theorem [latex](a^2 = b^2 + c^2, b^2 = a^2 + c^2, c^2 = a^2 + b^2)[\/latex] if [latex]A, B[\/latex] or [latex]C[\/latex] is 90\u00b0, since [latex]\\text{cos }90^{\\circ} = 0.[\/latex] The following examples illustrate the usage of the cosine law in trigonometry<\/p>\n

    \n
    \n

    Example 11.8.5<\/p>\n<\/header>\n

    \n

    Find the unknown angle shown in the triangle shown below.<\/p>\n

    \"Triangle<\/p>\n

    For this triangle:<\/p>\n

    [latex]\\begin{array}{l} A=\\text{find} \\\\ \\\\ a=90 \\\\ \\\\ b=50 \\\\ \\\\ c=100 \\end{array}[\/latex]<\/p>\n

    Note that [latex]b[\/latex] and [latex]c[\/latex] could be switched around. Now, to solve:<\/p>\n

    [latex]\\begin{array}{rrlllll} a^2 &= &\\phantom{-}b^2 &+ &c^2 &-& 2bc\\text{ cos }A \\\\ \\\\ 90^2 &= &\\phantom{-}50^2 &+ &100^2 &- &2(50)(100)\\text{ cos } A \\\\ 8100& =&\\phantom{-}2500 &+ &10000& - &10000\\text{ cos }A \\\\ -2500&&-2500&-&10000&& \\\\ \\hline -4400& =&-10000&\\text{ cos }&A&& \\\\ \\\\ \\text{ cos }A& =&\\dfrac{-4400}{-10000}&&&& \\\\ \\\\ B& =&\\text{cos}^{-1}0.44&&&& \\\\ \\\\ B&=&63.9^{\\circ}&&&& \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 11.8.6<\/p>\n<\/header>\n

    \n

    Find the unknown side shown in the triangle shown below.<\/p>\n

    \"TriangleFor this triangle:<\/p>\n

    [latex]\\begin{array}{l} B=70^{\\circ} \\\\ \\\\ a=95 \\\\ \\\\ b=\\text{find} \\\\ \\\\ c=60 \\end{array}[\/latex]<\/p>\n

    Note that [latex]b[\/latex] and [latex]c[\/latex] could be switched around. Now, to solve:<\/p>\n

    [latex]\\begin{array}{l} b^2=a^2+c^2-2ac\\text{ cos }B \\\\ \\\\ b^2=95^2+60^2-2(95)(60)\\text{ cos }70^{\\circ} \\\\ b^2=9025+3600-11400(0.34202) \\\\ b^2=12625 - 3899 \\\\ b^2=8726 \\\\ \\\\ b=\\sqrt{8726} \\\\ b=93.4 \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n

    Unlike with the law of sines, there should be no ambiguous cases with the law of cosines.<\/p>\n

    Questions<\/h1>\n

    Solve all unknowns in the following non-right triangles using either the law of sines or cosines.<\/p>\n

      \n
    1. \"Traingle<\/li>\n
    2. \"Triangle<\/li>\n
    3. \"Triangle<\/li>\n
    4. \"Triangle<\/li>\n
    5. \"Traingle<\/li>\n
    6. \"Traingle<\/li>\n
    7. \"Triangle<\/li>\n
    8. \"Triangle<\/li>\n
    9. \"Traingle<\/li>\n
    10. \"Triangle<\/li>\n
    11. \"Triangle<\/li>\n
    12. \"Triangle<\/li>\n
    13. \"Triangle<\/li>\n
    14. \"Triangle<\/li>\n
    15. \"Triangle<\/li>\n
    16. \"Triangle<\/li>\n<\/ol>\n

      Answer Key 11.8<\/a><\/p>\n","protected":false},"author":90,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1562","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1491,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1562","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1562\/revisions"}],"predecessor-version":[{"id":2180,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1562\/revisions\/2180"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1491"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1562\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1562"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1562"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1562"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}