In the triangle $ABC$, $AB=x$, $BC=y$, $AC=\frac{1}{6}$, angle $ABC=\frac{\pi }{6}$ radians and angl

Question

In the triangle $ABC$, $AB=x$, $BC=y$, $AC=\frac{1}{6}$, angle $ABC=\frac{\pi }{6}$ radians and angle $ACB=	heta $ radians (see diagram). Show that $\frac{x}{y}=\frac{2\sin 	heta }{\cos 	heta +\sqrt{3}\sin 	heta }$. [3] Given that $	heta $ is sufficiently small, express $\frac{x}{y}$ as a cubic polynomial in $	heta $. [3] Show that $	heta ={{\sin }^{-1}}\left( 3x ight)$. [1] Find the Maclaurin series for $	heta $, up to and including the term in ${{x}^{3}}$. [5]

Accepted Answer

$\dfrac{x}{y} \approx 2\theta - 2\sqrt{3}\,\theta^2 + \dfrac{20}{3}\theta^3$

2016 HCI P1 Q10

2016 HCI P1 Q10

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