SOLUTION: the field of view for a camera with a 200 millimeter lens is 12 degrees. A photographer takes a photograph of a large building that is 485 feet in front of the camera. What is the
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Question 890590: the field of view for a camera with a 200 millimeter lens is 12 degrees. A photographer takes a photograph of a large building that is 485 feet in front of the camera. What is the approximate width, to the nearest foot, of the building that will appear in the photograph? what is the answer Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! if i understand this correctly, the field of view of 12 degrees creates an isosceles triangle with a vertex of 12 degrees and an altitude of 485 feet.
the altitude to this isosceles triangle breaks it up into 2 right triangles, each with a vertex of 6 degrees.
the leg opposite these 6 degree angles divided by the altitude of these right triangle is equal to the tangent of 6 degrees.
you get:
tan(6) = opposite / altitude = opposite / 485.
solve for the opposite side to get:
opposite side = 485 * tangent of 6 degrees which is equal to 50.9755... feet.
the base of the isosceles triangle is 2 times that, so the base of the isosceles triangle is equal to 101.9511... feet.
the width of the building that will appear in the photograph is at most 102 feet rounded to the nearest foot.
here's a picture of what i think is happening.
this would be a top view of the camera and the building.
