If we let \(\beta\) be the angle between the forces 10N and 16N,
then \(\beta\) = (180\(^o\) - \(\theta\))
Applying the cosine rule; cos\(\theta\) = \(\frac{10^2 + 16^2 - 14^2}{2 \times 10 \times 16}\) = 0.5
If \(\theta\) = cos\(^{-1}\)(0.5) = 60\(^o\)
\(\beta\) = 180\(^o\) - 60\(^o\) = 120\(^o\)