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A picture 1.4 meters high stands on a wall so that its lower edge is 1.8 meters above the eye of an observer. What is the most favorable distance from the wall for this observer to stand (that is, to maximize his or her angle of vision)?
I've tried using the tan (pheta+beta) = (1.4+1.8)/x equation
but it still doesn't work out...
I've tried using the tan (pheta+beta) = (1.4+1.8)/x equation
but it still doesn't work out...