Consider the non-right triangle below 19610800141.jpg Suppose that m\angle BCA=69^{\circ}, and that

peromvu 2021-11-21 Answered
Consider the non-right triangle below
image
Suppose that \(\displaystyle{m}\angle{B}{C}{A}={69}^{{\circ}}\), and that \(\displaystyle{x}={32}{c}{m}\) and \(\displaystyle{y}={49}{c}{m}\). What is the degree measure of \(\displaystyle\angle{A}{B}{C}?\)

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Expert Answer

James Obrien
Answered 2021-11-22 Author has 1304 answers

Step 1
Use cosine formula
image
\(\displaystyle{x}={32},\ {y}={42}{c}{m}\)
\(\displaystyle\angle{B}{C}{A}={69}^{{\circ}}\)
\(\displaystyle\angle{A}{B}{C}=?\)
\(\displaystyle{{\cos{{69}}}^{{\circ}}\Rightarrow}{\frac{{{x}^{{{2}}}+{y}^{{{2}}}-{2}^{{{2}}}}}{{{2}{x}{y}}}}\)
\(3583=\frac{32^{2}+49^{2}\cdot2^{2}}{2\times32\times49}\)
\(\displaystyle{3563}\times{2}\times{32}\times{49}=\)
\(\displaystyle{1123}\cdot{8418}={1024}+{2401}-{2}^{{{2}}}\)
\(\displaystyle{1123}\cdot{8418}={3425}-{2}^{{{2}}}\)
\(\displaystyle{z}^{{{2}}}\Rightarrow{3425}-{1123}\cdot{8418}\)
\(\displaystyle{z}^{{{2}}}={2301}\cdot{1582}\)
\(\displaystyle{z}={47}\cdot{9703}\)
\(\displaystyle\angle{\cos{{B}}}={\frac{{{z}^{{{2}}}+{x}^{{{2}}}-{y}^{{{2}}}}}{{{22}{x}}}}\)
\(\displaystyle\angle{\cos{{B}}}={\frac{{{\left({47}\cdot{9703}\right)}^{{{2}}}+{32}^{{{2}}}-{49}^{{{2}}}}}{{{2}\times{47}\cdot{9703}\times{32}}}}\)
Step 2
\(\displaystyle{\cos{{B}}}={\frac{{{2301}\cdot{1582}+{1024}-{2401}}}{{{3070.0992}}}}\)
\(\displaystyle{\cos{{B}}}\Rightarrow{\frac{{{924.1582}}}{{{3070.0992}}}}\)
\(\displaystyle{\cos{{B}}}\Rightarrow\cdot{3010}\)
\(\displaystyle\angle{A}{B}{C}={B}={{\cos}^{{-{1}}}{\left(\cdot{3010}\right)}}\)
\(\displaystyle\angle{A}{B}{C}={72.4811}^{{\circ}}\)

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