A 1900kg car experiences a combinded force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27m/s. Going up a hill, the car's engine needs to produce 47 hp (1horsepower=745.7 Watts) more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

Well as far as understanding your diagram, I'm on the mark. But as for your equations, I'm lost lol. But thanks for the help, I'm sure once I get my sleep/early morning coffee I'll be able to figure something out.

as there are parallel and perpendicular lines everywhere (add in the horizontal and vertical at the point on the hill) the @ can be moved around with alternate and vertically opposite angles. from there the vectors are resolved into their horizontal and vertical components

to give you an understanding of how this works:

the ascii diagram didnt work so imagine a right triangle with hypotenuse v, base x and vertical side y with angle between hypotenuse and base = @

if the vector is v with angle @ then x and y are the horizontal and vertical components:

cos@ = x/v
x = v.cos@

sin@ = y/v
y = v.sin@

that is what i have done in the earlier explanation and all horizontal components have to cancel in either direction as to vertical components in order to have no acceleration

ok so far so good. Ok so, I don't have F, E and @ variables but I have mg which gives me N?

ok so F=E+X
F=mgsin@

so, mgsin@=E+1298. I can't solve for E, I don't have @.

Ok, so N=mgcos@ right. That doesn't make sense, if mg and N=mg, then wouldn't the values cancel out. err.

Ok, I've drawn something on your diagram, to see if I'm looking for the right angle. I used paint so it's a little "butchered" lol. I can see why this question was double starred for difficulty

you can't just say that N=mgcos@ because there are other vectors working in the vertical as well (parts of E & F)

i included 19000 (equivalent of mg) just to get rid of more variables so it's easier to solve for the others - it is only in the vertical components that it is included

i am so annoyed that i never saw that in the first place, you actually don't need to consider any of the other vectors besides mg as mg.sin@ is always the same, down the hill. so you can skip straight to 2mg.sin@=1298N

well at least you now have a mathematical proof of it