Ok, I'll give it a go. But this is way beyond anything I remember learning for the ATPL exams.
The first part is fairly simple. Yes, you do need to factor a vertical component of thrust, and a downward component of drag. The easiest way of doing this is to resolve all the forces relative to the aircraft (rather than the ground) - if you do it this way, T, L and D are all at right-angles to each other, and W is the only one that isn't. So you need to break W into the bit that oposes L, which is WcosA, and the bit that oposes T, which is WsinA, where A is the climb angle. All of this is fairly basic stuff, and is covered in PPL books such as Trevor Thom - although it makes a lot more sense with a diagram!
Now it starts getting more difficult.
We know that T=120kN.
We know that W=500kN.
We know that L=12D
The only force oposing L is the previously-calculated component of W:
L = WcosA = 500cosA
All the other forces are at right-angles to this one, along the thrust/drag line:
T = D + WsinA, or
120 = D + 500sinA. Re-arranging this gives us
D = 120 - 500sinA
Since L = 12D, and L = 500cosA, we know that:
12D = 500cosA, or
D = (500/12)cosA
We know have two formulae that give us D, so they must be equal:
120 - 500sinA = (500/12)cosA
Now, it's simply a case of solving this. But this is where my A-Level maths falls over, and I don't know where to go from here.
Fortunately, since it's a multi-choice question, we can simply put in each of the 4 values of A and see which one is the closest. The answers are given in terms of a gradient, so convert them to an angle by dividing by 100 then taking the inverse tangent. For answer b, 15.7% comes out as 8.9 degrees - and this fits the equation pretty much exactly. Answer = B.
I think.
Now I'm waiting for someone to come along with the easy method that I'd completely missed!
FFF
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