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## Homework Statement

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..

## Homework Equations

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 I am 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

[2bcsinA/((b^2+c^2-2bcCos(A))^1/2)] *δA

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.