(a)


Since the ladder is in equilibrium:
∑F = 0
∑Horizontal forces = 0
∑Vertical forces = 0
∑Torques about any point = 0 
Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— The linear momentum ()of its center of mass is zero.
— Its angular momentum ()about its center of mass, or about any other point, is also zero. 
Fg can be resolved into its horizontal and vertical components.
Fghorizontal = Fg Cos θ
Fgvertical = Fg Sin
θ 
Recall Sin θ = ^{opposite}/_{hypotenuse}
and Cos θ = ^{adjacent}/_{hypotenuse}
Fghorizontal is actually the friction from the ground
Fgvertical is the normal reaction of the floor 
Vertical forces:
[(W)] + Fgvertical = 0 Eq1
[(W)] + Fg Sin θ = 0
Fg Sin θ = W
Sin
θ = W/Fg
Sin
θ = 100/Fg  Eq2
Sin^{2}θ = [(100/Fg)]^{2}  Eq3 
Recall that we can treat with horizontal and vertical forces independently as a force has no component at right angles to it so vertical forces can have no horizontal forces and viceversa.
W = Fgvertical
Fgvertical is the normal reaction of the floor

Horizontal forces:
Fw + [Fghorizontal)] = 0 Eq4
Fw + [(Fg Cos θ)] = 0
Fg Cos θ = Fw
Cos
θ = Fw/Fg  Eq5
Cos^{2}θ = [(Fw/Fg)^{2}]  Eq6 
Fw = Fghorizontal
Fghorizontal is the friction from the floor 
Taking moments about B (anticlockwise is positive)
Fg X (0) + [(W X (½BC))] + [(Fw X (AC))] = 0
[(100N X (½ X 10 Cos 60°)] + [(Fw X 10 Sin 60°)] = 0
500 Cos 60° = 10 Fw
Sin 60°
Fw
Sin 60° = 50 Cos 60°
Fw
= (50 Cos 60°)/(Sin 60°)
Fw
= 50/(tan 60°)
Fw
= 28.9N 
Recall moment of a force = Force X perpendicular distance to force
Clockwise moments = Anticlockwise moments when in equilibrium.
We take moments through the point which Fg acts so its moment is zero.
Recall tan θ = ^{Sin θ}/_{Cos θ}

(b) Solving Fg and θ
Adding Eq3 and Eq6
Cos^{2}θ + Sin^{2}θ = [(100/Fg)]^{2} + [(Fw/Fg)^{2}]
1 = [(100/Fg)]^{2} + [(Fw/Fg)^{2}]
1 = 10000 / Fg^{2} + Fw^{2} / Fg^{2
}1 = (10000 + Fw^{2}) / Fg^{2}
Fg^{2} = 10000 + Fw^{2}
Fg^{2} = 10000 + 28.9^{2}
Fg = 10000 + 28.9^{2}
Fg = 104N
From Eq2:
θ = Sin^{1}(100/104)
From Eq5:
θ = Cos ^{1}(28.9/104)
θ = 74° 
Recall Cos^{2}θ + Sin^{2}θ = 1
Double Check:
Fg^{2} = Fgvertical^{2} + Fghorizontal^{2}
104^{2} = 28.9^{2} + 100^{2}
Also θ = tan ^{1}(100/28.9) 
(c) From Eq4:
Fghorizontal = Fw
Friction = 28.9N 
Double Check:
Fghorizontal = Fg Cos θ
Fghorizontal = 104 Cos 74°

(d) From Eq1:
Fgvertical = W
Normal Reaction from floor = 100N 
Double Check:
Fgvertical = Fg Sin θ
Fghorizontal = 104 Sin 74° 