Three forces 9, 12 and 15N acting at a point are in equilibrium. Then angle between 9N and 15N is

 $\sin {\left(9\right)}^2+{\left(12\right)}^2={\left(15\right)}^2$
 $Force9Nand12N$ are perpendicular.
 $Tan\alpha =\frac{12}{9}=\frac{4}{3}\sin \alpha =\frac{4}{5}$
 $\cot \alpha \frac{3}{5}$ 
Angle between 9N and 15N $\pi -\alpha =\pi -{\cos }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)$