Resolve horizontally.

The horizontal component of the \(20\text{ N}\) force is \(20\cos\theta\).

The horizontal component of the \(2F\text{ N}\) force is \(2F\cos30^\circ\).

So

\[ 20\cos\theta=2F\cos30^\circ. \]

Since \(\cos30^\circ=\frac{\sqrt3}{2}\),

\[ 20\cos\theta=F\sqrt3. \tag{1} \]

Resolve vertically.

The upward component of the \(20\text{ N}\) force is \(20\sin\theta\).

The downward forces are \(F\) and \(2F\sin30^\circ\).

So

\[ 20\sin\theta=F+2F\sin30^\circ. \]

Since \(\sin30^\circ=\frac12\),

\[ 20\sin\theta=F+F=2F. \]

Therefore

\[ F=10\sin\theta. \tag{2} \]

Substitute (2) into (1):

\[ 20\cos\theta=10\sqrt3\sin\theta. \]

So

\[ 2\cos\theta=\sqrt3\sin\theta. \]

Hence

\[ \tan\theta=\frac{2}{\sqrt3}. \]

Therefore

\[ \theta=49.1^\circ. \]

Using \(F=10\sin\theta\),

\[ F=10\sin49.1^\circ=7.56. \]

Answer: \(F=7.56\), \(\theta=49.1^\circ\).