Answer
When we try to use the law of cosines to find any angle A, B, or C, the angles are not defined. This makes sense since no triangle exists with the given values in the question.
Work Step by Step
We can try to use the law of cosines to find $A$:
$a^2 = b^2+c^2-2bc~cos~A$
$2bc~cos~A = b^2+c^2-a^2$
$cos~A = \frac{b^2+c^2-a^2}{2bc}$
$cos~A = \frac{4^2+10^2-3^2}{(2)(4)(10)}$
$cos~A = 1.3375$
Since there is no angle $A$ such that $cos~A \gt 1$, the angle $A$ is not defined.
We can try to use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$cos~B = \frac{3^2+10^2-4^2}{(2)(3)(10)}$
$cos~B = 1.55$
Since there is no angle $B$ such that $cos~B \gt 1$, the angle $B$ is not defined.
We can try to use the law of cosines to find $C$:
$c^2 = a^2+b^2-2ab~cos~C$
$2ab~cos~C = a^2+b^2-c^2$
$cos~C = \frac{a^2+b^2-c^2}{2ab}$
$cos~C = \frac{3^2+4^2-10^2}{(2)(3)(4)}$
$cos~C = -3.125$
Since there is no angle $C$ such that $cos~C \lt -1$, the angle $C$ is not defined.
