## Trigonometry (11th Edition) Clone

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.