al applied(friction)

Let d be the length of the ladder.

Point A, B and O be the point of contact of the ladder with the ground, the ladder with the wall and the wall with the ground respectively.

Let angle BAO = @ (theta), (I use $ as alpha)

Let the reaction force at A and B be R1 and R2 respectively,

Friction at A be F1.

Take horizontal equilibrium,

R2 = F1 cos$ - R1 sin$ ... (1)

Take vertical equilibrium,

R1 cos$ + F1 sin$ = W ... (2)

Take moment about point A,

W(d/2)cos(@ + $) = R2(d)sin(@ + $)

W = 2R2 tan(@ + $) ... (3)

Put (3) into (2): R1 cos$ + F1 sin$ = 2R2 tan(@ + $) ... (4)

Put (1) into (4): R1 cos$ + F1 sin$ = 2(F1 cos$ - R1 sin$) tan(@ + $)

R1 [cos$ + 2sin$ tan(@ + $)] = F1 [2cos$ tan(@ + $) - sin$] ... (5)

As the system is in limiting equilibrium,

tanx = F1 / R1 ... (6)

Comparing (5) and (6):

tanx = [cos$ + 2sin$ tan(@ + $)] / [2cos$ tan(@ + $) - sin$]

tanx [2cos$ tan(@ + $) - sin$] = [cos$ + 2sin$ tan(@ + $)]

tan(@ + $) [2tanx cos$ - 2sin$] = cos$ + tanx sin$

tan(@ + $) = [cos$ + tanx sin$] / [2tanx cos$ - 2sin$]

where 90* - (@ + $) is the inclination of the ladder to the wall.

so, tan[90* - (@ + $)] = [1 + tanx tan$] / 2[tanx - tan$]

tan(@ + $) = 2(tanx - tan$) / (1 + tanx tan$) = 2 tan(x - $)

the inclination, @ + $ = arctan [2 tan(x - $)]