Answer: 13.1 degrees
Step-by-step explanation:
Let N be the normal reaction of the floor on the ladder
Let R be the normal reaction of the wall on the ladder.
Let W be the weight of the (assumed uniform) ladder.
Let L be the length of the ladder
Summing horizontal forces to zero
0.115N = R
N = R / 0.115
Summing vertical forces to zero
N + 0.253 R = W
R / 0.115 + 0.253R = W
Sum moments about the floor contact to zero
W[(L/2)sinθ] - R[Lcosθ] - 0.253R[Lsinθ] = 0
Divide all terms by L and sub for W
(R / 0.115 + 0.253R)[(1/2)sinθ] - R[cosθ] - 0.253R[sinθ] = 0
expand
(Rsinθ / (2)0.115 + (0.253Rsinθ)/2) - R[cosθ] - 0.253R[sinθ] = 0
Divide all terms by R and combine
sinθ / 0.23 + 0.1265sinθ - cosθ - 0.253sinθ = 0
sinθ / 0.23 + 0.1265sinθ - 0.253sinθ = cosθ
4.2393sinθ = cosθ
tanθ = 1 / 4.2393
θ = 13.3 degrees