Visitors enter the museum through an enormous glass entryway in the shape of a tetrahedron. the figure shows the dimensions of the tetrahedron. the pitch is angle θ. what is the pitch of the tetrahedron’s slanted façade to the nearest degree? (hint: first find the sides of the big right isosceles triangle with a base of 260 feet. then, look at the smaller green triangle to find θ using the inverse sine.)

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°

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