calculus

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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

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