Notes: `a = 75.4, b = 52, c = 52` Use the Law of Cosines to solve the triangle. Round your answers to two decimal places.

Given: $\displaystyle{a}={75.4},{b}={52},{c}={52}$

$cos(A)={b^2+c^2-a^2]/(2bc)$

$\displaystyle{\cos{{\left({A}\right)}}}=\frac{{{{52}}^{{2}}+{{52}}^{{2}}-{{75}}^{{2}}}}{{{2}\cdot{52}\cdot{52}}}$

$\displaystyle{\cos{{\left({A}\right)}}}=-{.04012}$

$\displaystyle{A}={\arccos{{\left(-{.04012}\right)}}}$

$\displaystyle{A}={{92.30}}^{\circ}$

$\displaystyle{\cos{{\left({B}\right)}}}=\frac{{{{a}}^{{2}}+{{c}}^{{2}}-{{b}}^{{2}}}}{{{2}{a}{c}}}$

$\displaystyle{\cos{{\left({B}\right)}}}=\frac{{{{75}}^{{2}}+{{52}}^{{2}}-{{52}}^{{2}}}}{{{2}\cdot{75}\cdot{52}}}$

$\displaystyle{\cos{{\left({B}\right)}}}={.7211}$

$\displaystyle{B}={\arccos{{\left({.7211}\right)}}}$

$\displaystyle{B}={{43.85}}^{\circ}$

$\displaystyle{C}={180}-{92.30}-{43.85}$

$\displaystyle{C}={{43.85}}^{\circ}$