Answer
$A=22.3^{\circ}$ and $B=108.2^{\circ}$ and $C=49.5^{\circ}$
Work Step by Step
Need to apply the law of cosines to find $B$ when the sides of a triangle are given.
$b^2 = a^2+c^2-2ac \ cos \ B$
This implies that $B = arccos(\dfrac{a^2+c^2-b^2}{2ac}) = arccos(\dfrac{4^2+8^2-10^2}{(2)(4)(8)})= arccos(\dfrac{-5}{16})= 108.2^{\circ}$
Apply the law of cosines to find $C$ when the sides of a triangle are given.
$c^2 = a^2+b^2-2ab \ cos \ C$
This implies that $C = arccos(\dfrac{a^2+b^2-c^2}{2ab}) = arccos(\dfrac{4^2+10^2-8^2}{(2)(4)(10)})= arccos(\dfrac{13}{20})= 49.5^{\circ}$
Now, find angle $A$.
$A+B+C = 180^{\circ} $
or, $A= 180^{\circ}-B-C\\= 180^{\circ}-108.2^{\circ}-49.5^{\circ}\\
= 22.3^{\circ}$
