Question

a security camera is mounted 3.2 m high. the angle of depression to a package on the floor is 53°. how far is the package from the wall under the camera?

Explanation:

Step1: Determine the angle of elevation

The angle of depression is \(53^{\circ}\), so the angle of elevation from the package to the camera is also \(53^{\circ}\) (alternate - interior angles). Let the height of the camera \(h = 3.2\) m and the distance of the package from the wall be \(x\).

Step2: Use the tangent function

We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). In a right - triangle formed by the camera, the point on the wall directly below the camera, and the package, \(\theta = 53^{\circ}\), the opposite side is the height of the camera \(h = 3.2\) m, and the adjacent side is the distance \(x\) we want to find.
Since \(\tan\theta=\frac{h}{x}\), and \(\theta = 53^{\circ}\), \(h = 3.2\) m. We know that \(\tan53^{\circ}\approx1.33\).
So, \(x=\frac{h}{\tan\theta}\).
Substitute \(h = 3.2\) m and \(\tan\theta=\tan53^{\circ}\approx1.33\) into the formula:
\(x=\frac{3.2}{1.33}\approx2.4\) m

Answer:

The package is approximately \(2.4\) m from the wall under the camera.