Two Sides of a triangular plate are measured as 125mm and 160mm, each to the nearest millimetre. The included angle is quoted as 60+-1. Calculate the length of the remaining side and the maximum possible error in this result
My first question is to the nearest millimetre? What does this mean, i think +-0.5 but when i think about this if it was 0.5 biggest it would be rounded up..
δu = Fxδx + Fyδy + Fzδz
The Attempt at a Solution
So firstly to find the length of the other side i used the cosine rule
a^2 = b^2 + c^2 - 2bccos(A)
Where b = 125
c = 160
A = pi/3
Which comes out as 145.7 which is correct.
The next stage i guess is to take partial derivatives.
My equation for a is a^2 so i assume i have to square root it so i can find δa im not sure but i guess i have to??
δa = partial with respect to b * δb + partial with respect to c *δc + partial with respect to A δA
Therefore a = (b^2 + c^2 - 2bcCos(A))^1/2
Okay this is where it gets messy.
Partial with respect to b:
[1/2 * 2b-2cCosA/(b^2+c^2-2bcCos(A))^1/2] *δb
Partial with respect to C
[1/2 * 2c - 2bcos(A)/((b^2+c^2-2bcCos(A))^1/2) ]*δc
Partial with respect to A
When i input values of δb and δc = +-0.5 and δA = +-1 i don't seem to get the right answer which is +- 2.6, i really don't see where I'm going wrong, as i did the similar method with a previous exercise.