#### Answer

$\theta = 14.25^{\circ}$

#### Work Step by Step

Let $a$ be the length of the line from the origin to the point (8,6).
$a = \sqrt{8^2+6^2} = 10$
Let $b$ be the length of the line from the origin to the point (12,5).
$b = \sqrt{12^2+5^2} = 13$
Let $c$ be the length of the line from the point (8,6) to the point (12,5).
$c = \sqrt{(12-8)^2+(5-6)^2} = \sqrt{17}$
We can use the law of cosines to find $\theta$:
$c^2 = a^2+b^2-2ab~cos~\theta$
$2ab~cos~\theta = a^2+b^2-c^2$
$cos~\theta = \frac{a^2+b^2-c^2}{2ab}$
$\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$
$\theta = arccos(\frac{10^2+13^2-\sqrt{17}^2}{(2)(10)(13)})$
$\theta = arccos(0.96923)$
$\theta = 14.25^{\circ}$