Problem 515 | Friction

$f_1 = 0.25(120 \cos \theta) = 30 \cos \theta$
 

$N_2 = N_1 + 200 \cos \theta$

$N_2 = 120 \cos \theta + 200 \cos \theta$

$N_2 = 320 \cos \theta$
 

$f_2 = 0.25(320 \cos \theta) = 80 \cos \theta$
 

$f_1 + f_2 = 200 \sin \theta$

$30 \cos \theta + 80 \cos \theta = 200 \sin \theta$

$110 \cos \theta = 200 \sin \theta$

$\dfrac{100}{200} = \dfrac{\sin \theta}{\cos \theta}$

$\tan \theta = \dfrac{11}{20}$

$\theta = 28.81^\circ$           answer