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anonymous
 one year ago
help me on this please..
The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the
most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at
what distance from the wall should the observer stand?
anonymous
 one year ago
help me on this please.. The lower edge of a mural, 12 ft high, is 6ft above an observer’s eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is maximum, at what distance from the wall should the observer stand?

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433506316981:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0tan X = 18/x where x is distance of the eye from the wall tan Y = 6/x so tan X = 3 tan Y tan ( X  Y) has to be a maximum

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0I can't think of an analytical way to do this but using the fact that tan X = 3 tan Y and a trial and error numerical method I get X = 60 and Y = 30 so the angle subtended by the mural = 30 degrees and the distance from the eye to the wall is easy to calculate tan 30 = 6 / x x = 6 / tan 30

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0this is at best an approximate answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand how u get x=60 and y=30?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0tan X = 3 tan Y i used trial and error i tried X = 55 so tan 55 = 3 tan Y this gave Y = 25.45 and X_Y = 29.54 X = 59 this gave X  Y= 29.1 X = 61 gave X  Y = 29.98 X = 60 gave X  Y= 30 so X  Y seems to peak at 30 degrees
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