Question

In triangle ABC, a=15, b=14, c=10.Find m<B

Trigonometry
ANSWERED
asked 2021-08-20

In triangle ABC, a=15, b=14, c=10. Find \(\displaystyle{m}{<}{B}\)

Expert Answers (1)

2021-08-21

The given case is SSS (three sides) so use the Law of Cosines: \(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{c}{\mathcal{{o}}}{s}{B}\)
Solve for B.
\(\displaystyle{2}{c}{\mathcal{{o}}}{s}{B}={a}^{{2}}+{c}^{{2}}-{b}^{{2}}\)
\(\displaystyle{\cos{{B}}}=\frac{{{a}^{{2}}+{c}^{{2}}-{b}^{{2}}}}{{2}}{a}{c}\)
\(\displaystyle{m}{<}{B}={\left({\cos}^{{-{{1}}}}\right)}{\left(\frac{{{a}^{{2}}+{c}^{{2}}-{b}^{{2}}}}{{2}}{a}{c}\right).}\)
\(\displaystyle{m}{<}{B}={\left({\cos}^{{-{{1}}}}\right)}{\left(\frac{{{15}^{{2}}+{10}^{{2}}-{14}^{{2}}}}{{2}}\right){\left({15}\right)}{\left({10}\right)}.}\)
\(\displaystyle{m}{<}{B}={{\cos}^{{-{{1}}}}{\left(\frac{{129}}{{300}}\right)}}\)
Use a calculator in DEGREE mode: \(\displaystyle{m}∠{B}≈{64.5}°\)

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