# Use the Law of Cosines to solve the triangles. Round

Use the Law of Cosines to solve the triangles. Round lengths to the nearest tenth and angle measures to the nearest degree.
$$\displaystyle{a}={5},\ {b}={5},\ {c}={5}$$
$$\displaystyle{a}={66},\ {b}={25},\ {c}={45}$$

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Alrew1959
Step 1
The law of cosines is given by
$$\displaystyle{\cos{{A}}}={\frac{{{b}^{{{2}}}+{c}^{{{2}}}-{a}^{{{2}}}}}{{{2}{b}{c}}}}$$
$$\displaystyle{\cos{{B}}}={\frac{{{a}^{{{2}}}+{c}^{{{2}}}-{b}^{{{2}}}}}{{{2}{a}{c}}}}$$
$$\displaystyle{\cos{{C}}}={\frac{{{a}^{{{2}}}+{b}^{{{2}}}-{c}^{{{2}}}}}{{{2}{a}{b}}}}$$
Step 2
Given
$$\displaystyle{a}={5},\ {b}={5},\ {c}={5}$$
Using law of cosines,
$$\displaystyle{\cos{{A}}}={\frac{{{\left({5}\right)}^{{{2}}}+{\left({5}\right)}^{{{2}}}-{\left({5}\right)}^{{{2}}}}}{{{2}{\left({5}\right)}{\left({5}\right)}}}}$$
$$\displaystyle{\cos{{A}}}={\frac{{{25}+{25}-{25}}}{{{50}}}}$$
$$\displaystyle{\cos{{A}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle{A}={{\cos}^{{-{1}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}$$
$$\displaystyle\Rightarrow{A}={60}^{{\circ}}$$
$$\displaystyle{\cos{{B}}}={\frac{{{\left({5}\right)}^{{{2}}}+{\left({5}\right)}^{{{2}}}-{\left({5}\right)}^{{{2}}}}}{{{2}{\left({5}\right)}{\left({5}\right)}}}}$$
$$\displaystyle{\cos{{B}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{B}={60}^{{\circ}}$$
$$\displaystyle{\cos{{C}}}={\frac{{{\left({5}\right)}^{{{2}}}+{\left({5}\right)}^{{{2}}}-{\left({5}\right)}^{{{2}}}}}{{{2}{\left({5}\right)}{\left({5}\right)}}}}$$
$$\displaystyle{\cos{{C}}}={\frac{{{1}}}{{{2}}}}$$
$$\displaystyle\Rightarrow{C}={60}^{{\circ}}$$
Therefore,
$$\displaystyle{A}={60}^{{\circ}}$$
$$\displaystyle{B}={60}^{{\circ}}$$
$$\displaystyle{C}={60}^{{\circ}}$$
Step 3
Given,
$$\displaystyle{a}={66},\ {b}={25},\ {c}={45}$$
Using law of cosines,
$$\displaystyle{\cos{{A}}}={\frac{{{\left({25}\right)}^{{{2}}}+{\left({45}\right)}^{{{2}}}-{\left({66}\right)}^{{{2}}}}}{{{2}{\left({25}\right)}{\left({45}\right)}}}}$$
$$\displaystyle{\cos{{A}}}={\frac{{{625}+{2025}-{4356}}}{{{2250}}}}$$
$$\displaystyle{\cos{{A}}}=-{\frac{{{853}}}{{{1125}}}}$$
$$\displaystyle{A}={{\cos}^{{-{1}}}{\left(-{\frac{{{853}}}{{{1125}}}}\right)}}$$
$$\displaystyle\Rightarrow{A}\approx{139}^{{\circ}}$$
$$\displaystyle{\cos{{B}}}={\frac{{{\left({66}\right)}^{{{2}}}+{\left({45}\right)}^{{{2}}}-{\left({25}\right)}^{{{2}}}}}{{{2}{\left({66}\right)}{\left({45}\right)}}}}$$
$$\displaystyle{\cos{{B}}}={\frac{{{1439}}}{{{1485}}}}$$
$$\displaystyle\Rightarrow{B}\approx{14}^{{\circ}}$$
$$\displaystyle{\cos{{C}}}={\frac{{{\left({66}\right)}^{{{2}}}+{\left({25}\right)}^{{{2}}}-{\left({45}\right)}^{{{2}}}}}{{{2}{\left({66}\right)}{\left({25}\right)}}}}$$
$$\displaystyle{\cos{{C}}}={\frac{{{739}}}{{{825}}}}$$
$$\displaystyle\Rightarrow{C}\approx{26}^{{\circ}}$$
Therefore,
$$\displaystyle{A}\approx{139}^{{\circ}}$$
$$\displaystyle{B}\approx{14}^{{\circ}}$$
$$\displaystyle{C}\approx{26}^{{\circ}}$$