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In a bank, 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?

User Zerovector
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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:


\tan (\theta)=(10)/(45)

Let's solve for theta


\begin{gathered} \theta=\arctan \mleft((10)/(45)\mright) \\ \theta=0.21866=12.53^(\circ\: ) \end{gathered}

In a bank, a security camera is placed 10 feet high on a wall that is 45 feet from-example-1
User Gokublack
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