You have a datum axis where things are measured from. On a homebuilt this is typically a longeron in the fuselage or it could be some other more “abstract” line (which makes it more difficult).
You have to remember a C-172 has a stall speed of 45-50 kt and a top speed of 120. An RV stalls at 40 kt and has a top speed of 180. 4.5 degrees sound a lot to me. On a super efficient airliner wing cruising at 30-40k alt this could be right, or a glider perhaps (if you can talk about cruise in a glider), but on a light aircraft this sounds like way outside the low drag zone of the wing.
Anyway, lets assume a C-172 cruises at 100 kt with 4.5 degree aoa. For simplicity an RV does the same 100 kts with 4.5 degree (in reality it would be less due to less wing loading). As you know, Lift, L = Cl*1/2*rho*S*v² With everything else constant, the RV increases to 150 kts, a nice cruise speed for an RV. The v² term has increased by a factor 1.5² = 2.25. For the lift, L, to remain constant, the Cl will need to be reduced by the same amount. Lets say Cl originally was 0.5, the the new Cl will be 0.5/2.25 = 0.22 Now google a “typical” lift vs alpha curve, and you will find that alpha is reduced by approximately 3 degrees.
Or simply use the relation Cl = 2*pi*alpha. Delta Cl = 0.27 which means delta alpha = 0.22/(2*pi) = 0.0443 rad = 2.53 degrees. So the 4.5 degree is reduced to 2 degrees. If the engine had a 2 degree nose down attitude, then increasing to full power will obviously make the engine point down due to the increased speed. Remaining at full power and lifting the nose until the engine thrust is vertical will cause the airplane to go up. Even with 4.5 degree aoa, this is a plausible situation
But, this 4.5 degree obviously is erroneous. The wing of the Cessna produces lots of lift also at zero degree aoa, and I cant imagine the designer would be happy with 4.5 degree in cruise and all the drag associated with it. From wind tunnel test of a C-172 the Cl vs alpha has been found:
From wikipedia the wing area, S = 16.2 m². Lets say m=1000kg (two persons), v = 100 kt (50 ms) and rho = 1.4 for air. Using the same formula for lift, only solving for Cl; Cl = L/(1/2*rho*S*v²) = 0.35 Looking at the picture Cl=0.35 corresponds almost exactly to alpha = 0. A C-172 cruises at zero aoa of the wing. This of course means that even in a C-172, lightly loaded and full of power, it will happily climb in a steady fashion and the wings produces MORE lift than the weight of the aircraft.
Now, can we kill that question, please 
LeSving wrote:
Now, can we kill that question, please
Again, I think the question should have been rephrased to mention stall speed rather than lift. That would make it independent of how the engine is aligned and it would also be more relevant for actual flying.
LeSving wrote:
This of course means that even in a C-172, lightly loaded and full of power, it will happily climb in a steady fashion and the wings produces MORE lift than the weight of the aircraft.Now, can we kill that question, please
Not so quick 
I wouldn’t draw too many conclusions from that wind tunnel diagram. There is others around that show an AoA of a few degrees during the cruise.
Further, realise that “power” is not in the lift equation. In the scenario that you paint, where the AoA supposedly is 0 degrees in the cruise. If the student then goes from ~65% to 100% power, nothing in the lift equation will change of Cl*1/2*rho*S*v². The only thing that will happen is that the nose will pitch up and the aircraft starts to climb in a steady climb. Lift will be equal to weight. Most likely the nose will now also point up, so lift will be slightly less than weight. Unless (and that is the trick that you’re unintentionally keeping out of view all the time), the student trims nose down, increasing the speed, and the nose points lower, to the horizon or even slightly below.
A bit further on RV’s, whilst your technical data may be correct, it appears most of them climb (even in an en-route climb) at a speed that is ~30-60 kts slower than what they cruise at. The only way to achieve this, again, is by applying nose up trim. Most efficient climb is ~80 kts slower than cruise, meaning even more nose up trim.
Archie wrote:
There is others around that show an AoA of a few degrees during the cruise.
Such as?
In absence of acceleration, the sum of forces have to be zero. That is all there is to it. Thrust is a force that can point in any arbitrary direction. If it happens to point up, then some other force have to be less than mg, or the airplane will start accelerating.
LeSving wrote:
If it happens to point up
And that’s the point. In 99.99% of the climbs thrust will point up, and Lift will be slightly less than weight. You are right. If lift wasn’t less than weight, the airplane would accelerate.
LeSving wrote:
Such as?
Archie wrote:
And that’s the point. In 99.99% of the climbs thrust will point up, and Lift will be slightly less than weight
Only when take off, climb, cruise, descend, land – is to be done at best efficiency (by some measure) seen separately. That is the problem with the question. It is purely a physics question, but it is bundled with untold/unwritten assumptions. The correct answer is dependent on those assumptions even though there are no right or wrong from a pure physics point of view. Why not reformulate the question to: "A C-172 is in a steady climb at Vx … " and maybe throw in a question or two about why climbing at Vx (and Vy) can potentially kill you, which is the only relevant reason to have this question in the first place.
Which reminds me of a fun and very informative article I read by John Deakin.
