# 10.2 Non-right triangles: law of cosines  (Page 2/8)

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Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.

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.
2. Apply the Law of Cosines to find the length of the unknown side or angle.
3. Apply the Law of Sines    or Cosines to find the measure of a second angle.
4. Compute the measure of the remaining angle.

## Finding the unknown side and angles of a sas triangle

Find the unknown side and angles of the triangle in [link] .

First, 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.

Each 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 $\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$ as we know the measurement of the opposite angle $\text{\hspace{0.17em}}\beta .$

Because we are solving for a length, we use only the positive square root. Now that we know the length $\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$ we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle $\text{\hspace{0.17em}}\alpha ,\text{\hspace{0.17em}}$ we have

The other possibility for $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ would be $\text{\hspace{0.17em}}\alpha =180°–56.3°\approx 123.7°.\text{\hspace{0.17em}}$ In the original diagram, $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is adjacent to the longest side, so $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is an acute angle and, therefore, $\text{\hspace{0.17em}}123.7°\text{\hspace{0.17em}}$ does not make sense. Notice that if we choose to apply the Law of Cosines    , we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between $\text{\hspace{0.17em}}0°\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}180°.\text{\hspace{0.17em}}$ Proceeding with $\text{\hspace{0.17em}}\alpha \approx 56.3°,\text{\hspace{0.17em}}$ we can then find the third angle of the triangle.

$\gamma =180°-30°-56.3°\approx 93.7°$

The complete set of angles and sides is

$\begin{array}{ll}\alpha \approx 56.3°\begin{array}{cccc}& & & \end{array}\hfill & a=10\hfill \\ \beta =30°\hfill & b\approx 6.013\hfill \\ \text{\hspace{0.17em}}\gamma \approx 93.7°\hfill & c=12\hfill \end{array}$

Find the missing side and angles of the given triangle: $\text{\hspace{0.17em}}\alpha =30°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=12,\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=24.$

$a\approx 14.9,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta \approx 23.8°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\gamma \approx 126.2°.$

## Solving for an angle of a sss triangle

Find the angle $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ for the given triangle if side $\text{\hspace{0.17em}}a=20,\text{\hspace{0.17em}}$ side $\text{\hspace{0.17em}}b=25,\text{\hspace{0.17em}}$ and side $\text{\hspace{0.17em}}c=18.$

For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle $\text{\hspace{0.17em}}\alpha ,\text{\hspace{0.17em}}$ we have

Given $\text{\hspace{0.17em}}a=5,b=7,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=10,\text{\hspace{0.17em}}$ find the missing angles.

$\alpha \approx 27.7°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta \approx 40.5°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\gamma \approx 111.8°$

## Solving applied problems using the law of cosines

Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.

answer and questions in exercise 11.2 sums
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