Problem source

The lengths of the sides $BC$, $CA$, $AB$ of the triangle
$ABC$  are denoted by $a$, $b$, $c$, respectively. Given that 
$$ 
b = 8+{\epsilon}_1, \,
c=3+{\epsilon}_2,\,
A=\tfrac{1}{3}\pi + {\epsilon}_3,
$$
where ${\epsilon}_1$, ${\epsilon}_2$, and $ {\epsilon}_3$ are small, show that
$a \approx 7 + {\eta}$, where 
${\eta}= {\left(13 \, {{\epsilon}_1}-2\,{\epsilon}_2
+ 24{\sqrt 3} \;{{\epsilon}_3}\right)}/14$.
Given now that 
$$
{\vert {\epsilon}_1} \vert 
\le 2 \times 10^{-3}, \ \ \
{\vert {\epsilon}_2} \vert \le 4\cdot 9\times 10^{-2}, \ \ \  
{\vert {\epsilon}_3} \vert \le  \sqrt3 \times 10^{-3},
$$
find the range of possible values of ${\eta}$.

Solution source

The cosine rule states that:

$a^2 = b^2 + c^2 - 2bc \cos (A)$

Therefore

\begin{align*}
a^2 &= (8 + \epsilon_1)^2 + (3 + \epsilon_2)^2 - 2(8 + \epsilon_1) (3 + \epsilon_2)\cos \l \frac{\pi}{3}  + \epsilon_3 \r \\
&\approx 64 + 16\epsilon_1 + 9 + 6\epsilon_2- 2(24 + 3\epsilon_1+8\epsilon_2) \cos \l \frac{\pi}{3} + \epsilon_3  \r \\
&= 73 + 16\epsilon_1+ 6\epsilon_2 - 2(24 + 3\epsilon_1+8\epsilon_2) \l \cos \l \frac{\pi}{3} \r \cos \epsilon_3 - \sin \l \frac{\pi}{3} \r \sin \epsilon_3 \r \\
&\approx 73 + 16\epsilon_1+ 6\epsilon_2 - (24 + 3 \epsilon_1+8\epsilon_2) + 24\sqrt{3}\epsilon_3 \\
&= 49 + 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3 \\
&=  7^2 + 2 \cdot 7 \cdot \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \\
&\approx \l 7 + \frac{13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14} \r^2
\end{align*}
In this approximation, we are ignoring all terms of order $2$, and using the approximations $\cos \varepsilon \approx 1, \sin \varepsilon \approx \varepsilon$

Therefore $a \approx 7 + \frac{ 13 \epsilon_1 - 2\epsilon_2+24\sqrt{3}\epsilon_3}{14}$.

$\eta$ is maximised if $\epsilon_1, \epsilon_3$ are and $\epsilon_2$ is minimized, ie:

\begin{align*}
\eta &\leq  \frac{13 \cdot 2 \cdot 10^{-3} - 2 \cdot 4.9 \cdot 10^{-2} + 24 \sqrt{3} \cdot \sqrt{3} \cdot 10^{-3}}{14} \\
&= 10^{-3} \cdot \frac{26 - 98 + 74}{14} \\
&= 10^{-3} \cdot \frac{1}{7}\end{align*}

Similarly, it is maximised when signs are reversed, ie:

$| \eta | \leq 10^{-3} \cdot \frac{1}{7}$