a security camera in a bank is mounted on the wall 6.2 ft above the level of the counter and 9.4ft behind the front edge of the counter. What is the angle of depression of the camera if it is to be aimed 2 ft beyond the front edge of the counter.
i understand the problem but im having trouble setting it up because the behind the counter and the beyon the front edge part is confusing me. Please help me set this up.
This trigonometry, not calculus.
The camera gets aimed at a point 2 + 9.4 = 11.4 ft ahead of the vertical wall, from a point 6.2 ft above. The "angle of depression" of the camera, measured below a horizontal line is
arctan 6.2/11.4 = 28.5 degrees
To set up the problem, you can imagine a right triangle formed by the camera, the front edge of the counter, and the point the camera is aimed at.
Let's label the relevant points:
- The camera is mounted on the wall.
- The front edge of the counter is directly below the camera.
- The camera is aimed at a point 2 ft beyond the front edge of the counter, which is 9.4 ft behind the front edge of the counter.
To find the angle of depression, we need to consider the height (6.2 ft) and the horizontal distance (11.4 ft) between the camera and the point it is aimed at.
The angle of depression can be calculated using the tangent function:
tan(angle of depression) = height / horizontal distance
Plugging in the values into the equation:
tan(angle of depression) = 6.2 / 11.4
To solve for the angle of depression, we can use the inverse tangent function (also known as arctan or atan):
angle of depression = arctan(6.2 / 11.4)
Evaluating this expression, you will get approximately 28.5 degrees.