Answer
Magnitude of the resultant force $F_R$ is $F\sqrt {2(1+\cos\theta)}$
The angle between $\vec F_R$ and $\vec F_1$ is $\frac{\theta}{2}$
Work Step by Step
Two forces $\vec F_1$ and $\vec F_2$ act on the screw eye. The lines of action of these two forces are at an angle $\theta$ apart. The magnitude of each force is $F_1 = F_2 = F$.
Let $\vec F_R$ be the resultant force and the angle between $\vec F_R$ and $\vec F_1$ is $\alpha$
Using the law of cosines in the triangle formed by $F_1$, $F_2$, and $F_R$, we get,
$F_R=\sqrt {F_1^2+F_2^2-2F_1F_2\cos(180^{\circ}-\theta)}$
or, $F_R=\sqrt {F^2+F^2+2\times F\times F\times\cos\theta}$
or, $F_R=\sqrt {2F^2+2F^2\cos\theta}$
or, $F_R=F\sqrt {2(1+\cos\theta)}$
Using the law of sines in the triangle formed by $F_1$, $F_2$, and $F_R$, we get,
$\frac{F_1}{\sin(\theta-\alpha)}=\frac{F_2}{\sin\alpha}=\frac{F_R}{\sin(180^\circ-\theta)}$
$\therefore\frac{F_2}{\sin\alpha}=\frac{F_R}{\sin(180^\circ-\theta)}$
or, $\sin\alpha=\frac{F_2}{F_R}\times\sin(180^\circ-\theta)$
or, $\sin\alpha=\frac{F}{F\sqrt {2(1+\cos\theta)}}\times\sin\theta$
or, $\sin\alpha=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\sqrt {2\times2\cos^2\frac{\theta}{2}}}$
or, $\sin\alpha=\sin\frac{\theta}{2}$
or, $\alpha=\frac{\theta}{2}$
Therefore the magnitude of the resultant force $\vec F_R$ is $F\sqrt {2(1+\cos\theta)}$ and the angle between $\vec F_R$ and $\vec F_1$ is $\frac{\theta}{2}$
