Hi

The 20N force can be separated into 2 parts : 20 cos(theta)i+ 20 sin(theta)j

The projection on x axis (20 cos(theta)) gives no momentum with respect to H

The projection on y axis (20 sin(theta)) gives a momentum with respect to H equal to 20 sin(theta) x 5

The bar being at equilibrium, this momentum must be equal to the one generated by the 15N force, which is 15 x 1

20 sin(theta) x 5 = 15 x 1

sin(theta) = 3/20 = 0.15

theta = 8.6 deg