Trigonometry: Law of Cosines How to solve using rule of cosines? I can solve using law of sines but trying to check using rule of cosines is tripping me up, can anyone help clear things up?

 A: By writing down the law of cosines for this situation, we have
$$ 5^2=a^2+4^2-2\cdot a\cdot 4\cdot \cos 65^\circ$$
i.e.
$$ a^2-3.3809 a-9\approx 0.$$
That is a quadratic in $a$, hence you'll find two solutions, but one will be negative.
A: Get the angles as before:
$$65\,,\,46.47\,,\,68.53$$
and now the law of cosines:
$$a^2=5^2+4^2-2\cdot4\cdot5\cos 68.53^\circ=25.36\implies a=5.13$$
A: I'm new to using latex so please excuse the displayed equations. The underlying mathematics should be accurate. Feel free to correct it if you know how. 
The interior sum of angles of a triangle equal $180^\circ$. Therefore we can use [Law of Sines][1] to find $\angle A$ opposite a by deduction.
Let B be the angle opposite 4 and C=$65^\circ$. Then we have c=5.
$$\frac a{\sin A^ \circ}=\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac b{\sin B^\circ}=\frac c{\sin C^\circ}$$
$$\frac 4{\sin B^\circ}=\frac 5{\sin 65^\circ}$$
$$\frac 1{\sin B^\circ}=\frac 5{4\sin 65^\circ}$$
$${\sin B^\circ}=\frac {4\sin 65^\circ}5$$
$$B^\circ=\sin^{-1} ({\frac {4\sin 65^\circ}5})$$
$$B=46.47269373^\circ$$
Now we can find the third angle by [Sum of Angles of triangle][1] = $180^\circ$
$$180^\circ=A+B+C$$
$$180^\circ-(A+C)=B$$
$$B=68.53^\circ$$
Now you can use either [law of cosines] or [law of sines] to find $a$
Law of Cosines states:
$$a^2=b^2+c^2-2bc\cos A^\circ$$
$$a=\sqrt{b^2+c^2-2bc\cos A^\circ}$$
$$a=\sqrt{5^2+4^2-2(5x4)\cos 68.53^\circ}$$
$$a=4.91$$
