Law of Cosines

The Law of Cosines is the Pythagorean Theorem of all triangles, not just right triangles.

    \[a^2=b^2+c^2-2bc\cos(A)\]

    \[b^2=a^2+c^2-2ac\cos(B)\]

    \[c^2=a^2+b^2-2ab\cos(C)\]

Which one of these you use depends on what you’re trying to solve for.

Solving for a side: Side-Angle-Side

  1.  Label all of your points — This keeps them straight in your head.
  2. Plug the pieces into the proper places. b^2=a^2+c^2-2ac\cos(B)
  3. Evaluate.
  4. Take square root.

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    \[ \begin{array}{c c c l } \text{Step 2:}    & b^2     & =         & 5^2+3^2-2(5)(3)\cos(110)     \\        \vspace{12pt} & b^2     & =         & 25+9-30\cos(110)              \\     \vspace{12pt} \text{Step 3:}    & b^2     & \approx     & 44.2606043                 \\    %\vspace{12pt} \text{Step 4:}    & b     & =         & \sqrt{44.2606043} \approx 6.5  \\ \end{array} \]

Solving for an angle: Side – Side – Side

To solve for A, a^2=b^2+c^2-2bc\cos(A) could be used. Instead we will use an alternate version of the same equation. This is the same equation, except that we’re looking to solve and angle, so it’s automatically solved for the \cos(x). You can still use the above Law of Cosines to set up and solve.

    \[\cos(A)=\frac{a^2-b^2-c^2}{-2bc}\]

    \[\cos(B)=\frac{b^2-a^2-c^2}{-2ac}\]

    \[\cos(C)=\frac{c^2-a^2-b^2}{-2ab}\]

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\textbf{4 – step process to solve for an angle}

  1. Label all of your sides and angles.
  2. Plug your values into the equation.
  3. Evaluate.
  4. Calculate \cos^{-1} to determine the angle.

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    \[ \begin{array}{c c c l } \text{Step 2:}    & \cos(A)     & =         & \frac{8^2-5^2-9^2}{-2(5)(9)}     \\        \vspace{12pt} & \cos(A)     & =         & \frac{-42}{-90}              \\     \vspace{12pt} \text{Step 3:}    & \cos(A)     & \approx     & .466                 \\    %\vspace{12pt} \text{Step 4:}    & A             & =         & \cos^{-1}(.466) \approx 62  \\ \end{array} \]

 

Law of Cosines Proof

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    \[\cos(A)=\frac{x}{b} \Longrightarrow x=b\cos(A)\]

    \[h^2=b^2-x^2 \hspace*{1in} h^2=a^2-(c-x)^2\]

set h equal to itself

    \[b^2-x^2=a^2-(c-x)^2\]

expand the square of a binomial

    \[b^2-x^2=a^2-c^2+2cx-x^2\]

x^2 cancels out

    \[b^2=a^2-c^2+2cx\]

x=b\cos(A) from above

    \[b^2=a^2-c^2+2cb\cos(A)\]

solving for a^2

    \[a^2=b^2+c^2-2cb\cos(A)\]