Physics Problem?

I'll try the trajectory equation here: y = h + x·tanΦ - g·x² / (2v²·cos²Φ)
where y = height at x value of interest; here y = xtanΘ
and h = 0 m (assuming the ball is launched from the base of the incline)
and x = range value of interest = ???
and v = initial velocity = v0
and Φ = launch angle = Θ + φ
and Θ = angle of incline
and φ = angle of throw w/r/t incline

xtanΘ = 0 + xtan(Θ+φ) - gx² / (2(v0)²cos²(Θ+φ) → x cancels
tan(Θ+φ) - tanΘ = gx / (2(v0)²cos²(Θ+φ))
x = 2(v0)²cos²(Θ+φ)(tan(Θ+φ) - tanΘ) / g

The distance measured along the incline is
d = x / tanΘ = 2(v0)²cos²(Θ+φ)(tan(Θ+φ) - tanΘ) / gtanΘ
d = (2(v0)² / gtanΘ) * cos²(Θ+φ)(tan(Θ+φ) - tanΘ) ◄ distance

For the rest, I'll try using wolfram:
dd/dφ = 0 = secΘcos(Θ + 2φ)

I intepret the results to be
φ = π/4 - Θ/2 for the angles in radians, and
φ = 45º - Θ/2 for the angles in degrees.
Looks reasonable, I guess.

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