Question

So here's the question:

Answer 1

Answer:

The angle θ, that the device needs to scan is 16.795°

Step-by-step explanation:

Here we have

Height of room = 11 ft

Width of room = 15 ft

Length of room = 18 ft

Position of camera = 6 inches below the ceiling = 11 ft - 6 in = 126 in = 10.5 ft

Distance from camera to edge of long side of the room is given by the following relation;

Long edge of angle = √((18 ft)²+(10.5 ft)²) = 20.839 ft

Shorter edge of angle = √((15 ft)²+(10.5 ft)²) = 18.31 ft

Opposite side of required angle = √((18 ft)²+(15 ft)²) = 23.431 ft

Therefore, by cosine rule, we have

a² = b² + c² - 2·b·c·cos A

We therefore put our a as the opposite side of the required angle, A so we can easily solve for it

our b and c are then the other two sides

23.431² = 20.839² + 18.31² - 2×20.839×18.31×cosA

∴ cos(A) = (23.431² - (20.839² + 18.31²))÷(2×20.839×18.31)

cos(A) = 220.5/763.12418 = 0.29

A = cos⁻¹0.29 = 16.795°

The angle θ, that the device needs to scan = 16.795°.