So here's the question:
I am installing a camera in the corner of a living room 6 inches below the ceiling. I need to figure out the scan angle so that it can scan from one corner to the opposite side of the room. The room is 18 feet long, 15 feet wide, and 11 feet tall. I need to find the angle theta that the device needs to scan. There should also be a picture of it on here.
I was thinking I could find theta by looking at the top view of the box and finding it using the length and width measurements, but I feel like the height of the box has to be important, so I'm not sure. Any help would be appreciated.

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°.

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah