<p>In the figure below, $ABC$ is an equilateral triangle of length $2k$ units while $ABD$ is a triangle where&nbsp;$AD$ is of length $k$ units and $\measuredangle CAD$ is $x$ radians.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1759051530759-2s6cmhb6qet.png" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;$B{{D}^{2}}={{k}^{2}}\left( 5-2\cos x+2\sqrt{3}\sin x \right)$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$x$ is sufficiently small such that&nbsp;${{x}^{3}}$ and higher powers of $x$ may be may be neglected, find an approximation in $x$ for the length $BD$, leaving your answers in exact form and in terms of $k$.<span class="inline-marks">[2]</span></div></li></ol>

In the figure below, $ABC$ is an equilateral triangle of length $2k$ units while $ABD$ is a triangle where $AD$ is of length $k$ units and $\measuredangle CAD$ is $x$ radians.

Show that $B{{D}^{2}}={{k}^{2}}\left( 5-2\cos x+2\sqrt{3}\sin x \right)$.

Given that $x$ is sufficiently small such that ${{x}^{3}}$ and higher powers of $x$ may be may be neglected, find an approximation in $x$ for the length $BD$, leaving your answers in exact form and in terms of $k$.

2022 HCI P1 Q2

2022 HCI P1 Q2

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