Use the Law of Cosines to solve the triangles. Round

lenlifiauw2 2021-11-26 Answered
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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Expert Answer

Alrew1959
Answered 2021-11-27 Author has 6922 answers
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,
\(\displaystyle{A}\approx{139}^{{\circ}}\)
\(\displaystyle{B}\approx{14}^{{\circ}}\)
\(\displaystyle{C}\approx{26}^{{\circ}}\)
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