1 problem found
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: 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: \(| \eta | \leq 10^{-3} \cdot \frac{1}{7}\)