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estimate the maximum percentage error in Δ
...of a triangle, where Δ=[(bcsinA)/2] where b = 25 +/- 1c.m., c= 30 +/- 1.5cm,
A= 120+/-2degrees
δΔ = [(csinA)/2]δb + [(bsinA)/2]δc +[(bccosA)/2]δA
δΔ/Δ=δb/b + δc/c + cotAδΑ
so the percentage error is
(1/25)100 +(1.5/30)100 + [(cot120)(pi/180)2]100
= 4 +5 - 2 = 7% whereabouts but my book says its 11%
note I don't really understand that δA = (pi/180)2,
rather an exercises in my book computes in a similar context
δΑ =pi/180, where the error to angle A is +/- 1degree, so i just doubled it,
any tips?
A= 120+/-2degrees
δΔ = [(csinA)/2]δb + [(bsinA)/2]δc +[(bccosA)/2]δA
δΔ/Δ=δb/b + δc/c + cotAδΑ
so the percentage error is
(1/25)100 +(1.5/30)100 + [(cot120)(pi/180)2]100
= 4 +5 - 2 = 7% whereabouts but my book says its 11%
note I don't really understand that δA = (pi/180)2,
rather an exercises in my book computes in a similar context
δΑ =pi/180, where the error to angle A is +/- 1degree, so i just doubled it,
any tips?
Since the error can be plus or minus on each variable, I'd have though, for the maximum error, you wanted:
4 +5 + 2, which equals 11.
4 +5 + 2, which equals 11.
pb6883
though i am glad the right answer is found, i am not sure i follow...
cot120=-0.57??
cot120=-0.57??
The error in A is within Plus or Minus 2. If you use +2 then the overall error is less than if you use -2, and since we are looking for the maximum error....
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