Trigonometry: Law of Sines How can this be solved using law of sines? I get a different answer then when I solve it using law of cosines..
Let $\theta$ be the angle opposite $4$. Then we have 
$$\frac{\sin\theta}{4}=\frac{\sin(65^\circ)}{5}.$$
Calculation shows that $\theta$ is about $46.47$ degrees, or $180^\circ$ minus that. The second is impossible, it would make the angle sum greater than $180^\circ$.  

Now we can find the third angle by subtraction, and use the Sine Law again to find $a$.
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:
$$a^2=b^2+c^2-2bc\cos A^\circ$$
$$a=\sqrt{b^2+c^2-2bc\cos A^\circ}$$
$$a=\sqrt{5^2+4^2-2(5x4)\cos 68.53^\circ}$$
$$a=4.91$$