# Use the Law of Cosines to solve the triangles. Round lengths to the nearest tent

Use the Law of Cosines to solve the triangles. Round lengths to the nearest tenth and angle measures to the nearest degree.

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Step 1
Given that,
$$\displaystyle{A}={22}^{{\circ}},\ {b}={6},\ {c}={15}$$
using Cosine rule, $$\displaystyle{\cos{{A}}}={\frac{{{b}^{{{2}}}+{c}^{{{2}}}-{a}^{{{2}}}}}{{{2}{b}{c}}}}$$
$$\displaystyle\Rightarrow{\cos{{\left({22}\right)}}}={\frac{{{\left({6}\right)}^{{{2}}}+{\left({15}\right)}^{{{2}}}-{a}^{{{2}}}}}{{{2}{\left({6}\right)}{\left({15}\right)}}}}$$
$$\displaystyle\Rightarrow{0.9272}={\frac{{{36}+{225}-{a}^{{{2}}}}}{{{180}}}}$$
$$\displaystyle\Rightarrow{166.893}={261}-{a}^{{{2}}}$$
$$\displaystyle\Rightarrow{a}^{{{2}}}={94.1069}$$
$$\displaystyle\Rightarrow{a}={9.7009}$$
$$\displaystyle\Rightarrow{a}\approx{10}$$
Step 2
Using Sine rule, $$\displaystyle{\frac{{{\sin{{A}}}}}{{{a}}}}={\frac{{{\sin{{B}}}}}{{{b}}}}$$
$$\displaystyle\Rightarrow{\frac{{{\sin{{\left({22}\right)}}}}}{{{10}}}}={\frac{{{\sin{{B}}}}}{{{6}}}}$$
$$\displaystyle\Rightarrow{\sin{{B}}}={\frac{{{6}{\sin{{\left({22}\right)}}}}}{{{10}}}}$$
$$\displaystyle\Rightarrow{B}\approx{13}^{{\circ}}$$
using angle sum property of a triangle,
$$\displaystyle{C}={180}-{\left({A}+{B}\right)}$$
$$\displaystyle={180}-{\left({22}+{13}\right)}$$
$$\displaystyle={145}^{{\circ}}$$
Therefore, $$\displaystyle{a}={10},\ {B}={13}^{{\circ}},\ {C}={145}^{{\circ}}$$