a security camera is placed 10 feet high on a wall that is 45 feet from the front door
What angle of depression is needed so the camera aims for a point at the base of the front door?
Let's start by drawing a diagram of this situation
We can calculate the angle of depression using the following trigonometric function:
[tex]\tan (\theta)=\frac{10}{45}[/tex]Let's solve for theta
[tex]\begin{gathered} \theta=\arctan \mleft(\frac{10}{45}\mright) \\ \theta=0.21866=12.53^{\circ\: } \end{gathered}[/tex]
