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We have $\frac {\partial}{\partial x}\arctan (\frac xy)=\frac y{x^2+y^2}$ so the error in the angle due to error in $a$ is $\frac b{a^2+b^2}\Delta a$ You can do a similar thing for errors in $b$ and add them together to get the total error in $\theta$

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Three significant digit cancellation wont have an effect in this question and the answer will be same. The answer proposed in the question is the right one as 0.05/45.2 relative error.

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Three hints, depending on whether you like playing with numbers, algebra, or calculus: Numbers: Calculate the areas of circles with radii 12.1, 12.13, 12.07. Algebra: Expand $\pi(r+\varepsilon)^2$ and see how it differs from $\pi r^2$. Calculus: Ponder $\delta A \approx \tfrac{dA}{dr} \delta r$

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In general, the framework in which we work to estimate a parameter and to control the quality of the estimation is the following: we need a parameter $\vec \theta$ to be estimated, a set of observations $x$ and we look for a statistic $\hat\theta=\hat\theta(x)$ that estimates the parameter $\vec \theta$. The core of the "quality control" lies on the fact ...

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Error propagation is estimated by differentials, so letting $C = re^{i\phi}$, we have $dC = \frac{\partial C}{\partial r}dr = \frac{\partial C}{\partial \phi}d\phi = \frac {C}{r}dr + iCd\phi$. Or in your notation: $$\frac{\delta C}{C} = \frac {\delta |C|}{|C|} + i\delta \phi$$

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