In either case, the force applied is 110 N. That doesn't change. What changes then?

The problem tells you when the force applied is perfectly horizontal - the friction is "twice as great." But what is friction?

Friction = mk x Fnormal

Looking at this equation - you know that mk is a constant. Therefore, the only way the friction could be "twice as great" was if the normal force was twice as large.

So, if the normal force is twice as large (when the applied force is perfectly horizontal), it must be half as large when held at it's at an angle.

Let's consider only the vertical forces acting on the object:

You have the downward weight of the object.

You have the upward applied force.

You have the normal force (which we know needs to equal 75 N)

We also know the object isn't accelerating upwards or downwards, so Fnet = 0 N.

Let's set up the equation:

Fnet = 0N = Weight - Fapplied + Normal Force

Normal Force = Weight - Fapplied

75 N = 150 N - 110sin(theta) <== Note: "sin(theta)" because we're only considering the vertical component

110sin(theta) = 150N - 75N

110sin(theta) = 75N

sin(theta) = 75N/110N

sin(theta) = 0.68

So the angle "theta" must equal 42.98 or ~45 degrees if you had to estimate for the MCAT.