Suppose a TV with height of 60 feet has been mounted onto a platform so that the

Question

Suppose a TV with height of 60 feet has been mounted onto a platform so that the bottom of the TV is 150 feet off the ground. Viewing of the TV is optimum when ? has been maximized. Generate an array for distance from 60 to 350 feet at increments of 0.5 feet. Calculate ? in degrees for each of these distances. Determine the max value of theta and the position in the array where this value occurs. Use the position (index/subscript) of the maximum ? value and the array for distances to determine the best distance for viewing.

Explanation / Answer

$B$1 is 150, $C$1 is 210 60, =ATAN($B$1/A2), =ATAN($C$1/A2), =C2-B2 OK, first you would think that the closer you are the bigger the extended angle, but if you are right under the screen you are looking straight up to see the bottom and straight up to see the top, so the difference is zero (90-90) when you are 0 feet from the screen. If you are way far from the screen the top and bottom will appear very close, so this is a max/min problem. So the first line is how high up the bottom of the screen and the top of the screen are. The second line is the first position, 60 feet and I find the angle to the bottom of the screen; It is the inverse tangent of 150/60, and the angle to the top of the screen; it is the inverse tangent of 210/60. After finding the two angles, I subtract them. Then I increment the 60 feet by a half and keep going until I get to 350 feet. Actually you could stop once the value starts decreasing, but I kept going. The maximum angle occurs when the viewer is 177.5 feet from the point directly below the screen. 176.5 0.704411827 0.871857375 0.167445548 177 0.703016233 0.870463703 0.167447471 177.5 0.70162522 0.869073298 0.167448078 178 0.700238771 0.867686153 0.167447382 178.5 0.698856869 0.866302263 0.167445393 Theta for the max value is 0.167448078 radians or 9.594 degrees.

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