Question

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

Trigonometry

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

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: $$\displaystyle{m}∠{B}≈{64.5}°$$