The rest of the question is the attached figure.
Solution:
As shown in the figure
ΔABC is an isosceles right triangle at B
AB = BC and AC is the hypotenuse ⇒ AC = 260 ft
using Pythagorean theorem
∴ AC² = AB² + BC² = AB² + AB² = 2 AB²
∴ AB² = AC²/2 = 0.5 AC²
∴ AB = √(0.5 AC²) = √(0.5 * 260²) = 130√2 ft →(1)
Also as shown in the figure
ΔADB is a right triangle at D
from (1) AB = 130√2 ft
given BD = 105 ft
we should know that ⇒

∴ θ = sin⁻¹ 0.571 ≈ 34.82° = 35° ( to the nearest degree )
So, the value of θ = 35°