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 = (opposite)/(hypotenuse) = (BD)/(AB) = (105)/(130 √(2)) = 0.571`

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


So, the value of θ = 35°