Use the Law of Cosines to solve the triangles. Round

Lloyd Allen 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. PLEASE solve both of these triangles. Thank you!
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Expert Answer

Ched1950
Answered 2021-11-27 Author has 7966 answers
using Cosine rule, \(\displaystyle{\cos{{A}}}={\frac{{{b}^{{{2}}}+{c}^{{{2}}}-{a}^{{{2}}}}}{{{2}{b}{c}}}}\)
\(\displaystyle\Rightarrow{\cos{{\left({22}\right)}}}={\frac{{{\left({6}\right)}^{{{2}}}+{\left({15}\right)}^{{{2}}}-{a}^{{{2}}}}}{{{2}{\left({6}\right)}{\left({15}\right)}}}}\)
\(\displaystyle\Rightarrow\frac{{0}}{{9272}}={\frac{{{36}+{225}-{a}^{{{2}}}}}{{{180}}}}\)
\(\displaystyle\Rightarrow{166.893}={261}-{a}^{{{2}}}\)
\(\displaystyle\Rightarrow{a}^{{{2}}}={94.1069}\)
\(\displaystyle\Rightarrow{a}={9.7009}\)
\(\displaystyle\Rightarrow{a}\approx{10}\)
using Sine rule, \(\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}\)
\(\displaystyle\Rightarrow{\frac{{{\sin{{\left({22}\right)}}}}}{{{10}}}}={\frac{{{\sin{{B}}}}}{{{6}}}}\)
\(\displaystyle\Rightarrow{\sin{{B}}}={\frac{{{6}{\sin{{\left({22}\right)}}}}}{{{10}}}}\)
\(\displaystyle\Rightarrow{B}\approx{13}°\)
using angle sum property of a triangle,
\(\displaystyle{C}={180}-{\left({A}+{B}\right)}\)
\(\displaystyle={180}-{\left({22}+{13}\right)}\)
\(\displaystyle={145}^{{\circ}}\)
Therefore, \(\displaystyle{a}={10},{B}={13}^{{\circ}},{C}={145}^{{\circ}}\)
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