#### Answer

$\theta = 16.26^{\circ}$

#### Work Step by Step

Let $a$ be the length of the line from the origin to the point (6,8).
$a = \sqrt{6^2+8^2} = 10$
Let $b$ be the length of the line from the origin to the point (4,3).
$b = \sqrt{4^2+3^2} = 5$
Let $c$ be the length of the line from the point (4,3) to the point (6,8).
$c = \sqrt{(6-4)^2+(8-3)^2} = \sqrt{29}$
We can use the law of cosines to find $\theta$:
$c^2 = a^2+b^2-2ab~cos~\theta$
$2ab~cos~\theta = a^2+b^2-c^2$
$cos~\theta = \frac{a^2+b^2-c^2}{2ab}$
$\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$
$\theta = arccos(\frac{10^2+5^2-\sqrt{29}^2}{(2)(5)(10)})$
$\theta = arccos(0.96)$
$\theta = 16.26^{\circ}$