<p>In the triangle \(\Delta ABC\), \(AC = 1\), \(\angle BAC = \frac{\pi}{3}\) radians and \(\angle ABC = \left( \frac{\pi}{6} + \theta \right)\) radians.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;\(BC = \frac{\sqrt{3}}{\cos \theta + \sqrt{3} \sin \theta}\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;\(\theta\) is sufficiently small such that&nbsp;\(\theta^3\) and higher powers of \(\theta\) may be neglected, show that\[BC \approx \sqrt{3} (1 + a\theta + b\theta^2)\]where&nbsp;\(a\) and \(b\) are constants to be determined.<span class="inline-marks">[3]</span></div></li></ol>

In the triangle \(\Delta ABC\), \(AC = 1\), \(\angle BAC = \frac{\pi}{3}\) radians and \(\angle ABC = \left( \frac{\pi}{6} + \theta \right)\) radians.

Show that \(BC = \frac{\sqrt{3}}{\cos \theta + \sqrt{3} \sin \theta}\).

Given that \(\theta\) is sufficiently small such that \(\theta^3\) and higher powers of \(\theta\) may be neglected, show that\n\[BC \approx \sqrt{3} (1 + a\theta + b\theta^2)\]\nwhere \(a\) and \(b\) are constants to be determined.

Given that \(\theta\) is sufficiently small such that \(\theta^3\) and higher powers of \(\theta\) may be neglected, show that\[BC \approx \sqrt{3} (1 + a\theta + b\theta^2)\]where \(a\) and \(b\) are constants to be determined.

Maclaurin Series and Power Series

International Baccalaureate Higher Level Mathematics

2025 TJC P1 Q4

2025 TJC P1 Q4