Question

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.)

Answer 1

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 ⇒ $sin \ \theta = \frac{opposite}{hypotenuse} = \frac{BD}{AB} = \frac{105}{130 \sqrt{2}} = 0.571$

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


So, the value of θ = 35°