Skip to main content
K12 LibreTexts

4.1.8: Determination of Unknown Angles Using Law of Cosines

  • Page ID
    14872
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Find unknown angle given lengths of all 3 sides

    Sarine draws a triangle. She measures the length of the sides and records her measurements as follows. What is the measure of angle \(C\) of the triangle?

    \(\begin{aligned} a&=3 \\ b&=4 \\ c&=5 \end{aligned}\)

    Law of Cosines with SSS

    The Law of Cosines, \(a^2+b^2−2ab\cos C\), can be rearranged to facilitate the calculation of the measure of angle \(C\) when \(a\), \(b\) and \(c\) are all known lengths.

    \(\begin{aligned} a^2+b^2−2ab \cos C &=c^2 \\ a^2+b^2−c^2 &=2ab\cos C\\ \dfrac{a^2+b^2−c^2}{2ab} &=\cos C \end{aligned}\)

    which can be further manipulated to \(C=\cos ^{−1}\left(\dfrac{a^2+b^2−c^2 }{2ab}\right)\).

    Let's find the measure of the largest angle in the triangle with side lengths 12, 18 and 21.

    First, we must determine which angle will be the largest. Recall from Geometry that the longest side is opposite the largest angle. The longest side is 21 so we will let \(c=21\) since \(C\) is the angle we are trying to find. Let \(a=12\) and \(b=18\) and use the formula to solve for \(C\) as shown. It doesn’t matter which sides we assign to \(a\) and \(b\). They are interchangeable in the formula.

    \(m\angle C=\cos ^{−1}\left(\dfrac{12^2+18^2−21^2 }{2(12)(18)}\right) \approx 86^{\circ}\)

    Note: Be careful to put parenthesis around the entire numerator and entire denominator on the calculator to ensure the proper order of operations. Your calculator screen should look like this:

    \(\cos ^{−1}((12^2+18^2−21^2)/(2(12)(18)))\)

    Now let's find the value of \(x\), to the nearest degree.


    f-d_924c7ef78d4e2e2ee8bc5e85d1d29f9df5ff4c34db630b776b65b91e+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{1}\)

    The angle with measure \(x^{\circ} \) will be angle \(C\) so \(c=16\), \(a=22\) and \(b=8\). Remember, \(a\) and \(b\) are interchangeable in the formula. Now we can replace the variables with the known measures and solve.

    \(\cos ^{−1}\left(\dfrac{22^2+8^2−16^2}{2(22)(8)}\right)\approx 34^{\circ}\)

    Finally, let's find the \(m\angle A\), if \(a=10\), \(b=15\) and \(c=21\).

    First, let’s rearrange the formula to reflect the sides given and requested angle:

    \(\cos A=\left(\dfrac{b^2+c^2−a^2}{2(b)(c)}\right)\), now plug in our values

    \(m\angle A=\cos ^{−1}\left( \dfrac{15^2+21^2−10^2}{2(15)(21)}\right) \approx 26^{\circ}\)

    Example \(\PageIndex{1}\)

    Earlier, you were asked to find the measure of angle \(C\) of the triangle that has sides \(a = 3\), \(b = 4\), and \(c = 5\).

    Solution

    We can use the manipulated Law of Cosines to solve for \(C\).

    \(\begin{aligned} C&=\cos ^{−1} \dfrac{3^2+4^2−5^2}{2(3)(4)} \\ C&=\cos ^{−1} \dfrac{9+16−25}{24} \\ C&=\cos ^{−1} \dfrac{0}{24}=\cos ^{−1} 0 \\ C&=90^{\circ} \end{aligned}\)

    Therefore, the triangle is a right triangle.

    Example \(\PageIndex{2}\)

    Find the measure of \(x\) in the diagram:


    f-d_e07b58d594d1fffb34f9f8c99789065d7b5e9a9737067e125db8186f+IMAGE_TINY+IMAGE_TINY.png
    Figure \(\PageIndex{2}\)

    Solution

    \(\cos ^{−1}\left(\dfrac{14^2+8^2−19^2}{2(14)(8)}\right) \approx 117^{\circ}\)

    Example \(\PageIndex{3}\)

    Find the measure of the smallest angle in the triangle with side lengths 47, 54 and 72.

    Solution

    The smallest angle will be opposite the side with length 47, so this will be our \(c\) in the equation.

    \(\cos ^{−1}\left(\dfrac{54^2+72^2−47^2 }{2(54)(72)}\right) \approx 41^{\circ}\)

    Example \(\PageIndex{4}\)

    Find \(m\angle B\), if \(a=68\), \(b=56\) and \(c=25\).

    Solution

    Rearrange the formula to solve for \(m\angle B\),

    \(\cos B=\left(\dfrac{a^2+c^2−b^2}{2(a)(c)}\right)\); \(\cos ^{−1}\left(\dfrac{68^2+25^2−56^2}{2(68)(25)}\right)\approx 52^{\circ}\)

    Review

    Use the Law of Cosines to find the value of \(x\), to the nearest degree, in problems 1 through 6.


    1. f-d_119780a9b2c0a790886584910c1b4152641299c82f60c41f7b8f4360+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{3}\)

    2. f-d_d670828cfcb2d513d0e9dceffc5c7ec75db4d441071ec63b5a414e36+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{4}\)

    3. f-d_8cd33ac9b8e12114f01c56648e82708215121931cabec6a326867d07+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{5}\)

    4. f-d_1d1d5c91d583998e2b66575c32a17efd7a002e3f5358849ab81ab013+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{6}\)

    5. f-d_8e03abbab7ca6dea50739fa072b059081c59fbcc96a5570c41879d0f+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{7}\)

    6. f-d_beb382b2b7eac155726760888478f16c078a975395d0de38105d4640+IMAGE_TINY+IMAGE_TINY.png
      Figure \(\PageIndex{8}\)
    7. Find the measure of the smallest angle in the triangle with side lengths 150, 165 and 200 meters.
    8. Find the measure of the largest angle in the triangle with side length 59, 83 and 100 yards.
    9. Find the \(m\angle C\) if \(a=6\), \(b=9\) and \(c=13\).
    10. Find the \(m\angle B\) if \(a=15\), \(b=8\) and \(c=9\).
    11. Find the \(m\angle A\) if \(a=24\), \(b=20\) and \(c=14\).
    12. A triangular plot of land is bordered by a road, a fence and a creek. If the stretch along the road is 100 meters, the length of the fence is 115 meters and the side along the creek is 90 meters, at what angle do the fence and road meet?

    Answers for Review Problems

    To see the Review answers, open this PDF file and look for section 13.16.


    This page titled 4.1.8: Determination of Unknown Angles Using Law of Cosines is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.

    CK-12 Foundation
    LICENSED UNDER
    CK-12 Foundation is licensed under CK-12 Curriculum Materials License
    • Was this article helpful?