SOLUTION: A device used in golf to estimate the distance​ d, in​ yards, to a hole measures the size​ s, in​ inches, that the 8​-ft pin appears to be in a viewfi
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Question 1100073: A device used in golf to estimate the distance d, in yards, to a hole measures the size s, in inches, that the 8-ft pin appears to be in a viewfinder. The viewfinder is held 3 inches from the viewer's eye. Express the distance d as a function of s.
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
The 8 ft pin is 8/3 yards tall. We need to express the pin height in yards so the units will cancel when we take the ratio with distance 'd'.
Drawing a triangle from the viewers eye to the pin, up the pin, and back to the viewer's eye; and drawing a 2nd triangle from the viewers eye to the edge of the viewfinder up to the apparent height of the pin (the image) in the viewfinder and back to the viewer's eye gives two similar triangles.
The ratio of the height of the image in the viewfinder to its horizontal length (3in) is equal to the ratio of the pin height (8/3 yards) to the distance d to the pin, in yards. The units cancel in finding these ratios so we can write:
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s/3 = (8/3)/d
s*d = (8/3)*3
s*d = 8
Remembering that s is in inches and d is in yards.
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I think this is correct. Qualitatively, we can check that it at least makes sense: when the distance d is large, we expect s to be small (since far away the pin will look smaller). As one gets closer to the pin, the image s gets larger and d computes to a smaller value. Our formula has those characteristics.
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