In the triangle $ABC$, $AC=1$, angle $BAC=x$ radians and angle $ACB=\frac{1}{6}\pi $ radians

Question

In the triangle $ABC$, $AC=1$, angle $BAC=x$ radians and angle $ACB=\frac{1}{6}\pi $ radians (see diagram). Show that $AB=\frac{1}{\cos x+\sqrt{3}\sin x}$.[3] Given that $x$ is a sufficiently small angle, show that $AB\approx 1+ax+b{{x}^{2}}$, for constants $a$ and $b$ to be determined exactly.[3] The first two terms in the Maclaurin series for $\ln \left( px+q \right)$ are equal to the first two terms in the series expansion in part (b). Using standard series from the List of Formulae (MF26), find the exact values of $p$ and $q$. [4]

Accepted Answer

(a) $AB=\frac{1}{\cos x+\sqrt3\sin x}$ (b) $a=-\sqrt3$, $b=\frac72$ (c) $p=-\sqrt3\,\mathrm e$, $q=\mathrm e$

2023 YIJC P2 Q5

2023 YIJC P2 Q5

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