Answer
The angle between the forces is $93.9^{\circ}$
Work Step by Step
Let $a = 692~N$
Let $b = 423~N$
Let $c$ be the resultant force of $786~N$
Let $\theta$ be the angle between $a$ and $b$
We can use the parallelogram rule to find $\theta$:
$c^2 = a^2+b^2+2ab~cos~\theta$
$-2ab~cos~\theta = a^2+b^2-c^2$
$cos~\theta = -\frac{a^2+b^2-c^2}{2ab}$
$\theta = arccos(-\frac{a^2+b^2-c^2}{2ab})$
$\theta = arccos(-\frac{692^2+423^2-786^2}{(2)(692)(423)})$
$\theta = arccos(-0.06832)$
$\theta = 93.9^{\circ}$
The angle between the forces is $93.9^{\circ}$
