In the triangle $ABC$, $AB=1$, angle $BAC=\theta $ radians and angle $ABC=\frac{2\pi }{3}$ r

Question

In the triangle $ABC$, $AB=1$, angle $BAC=\theta $ radians and angle $ABC=\frac{2\pi }{3}$ radians (see diagram). Show that $AC=\frac{\sqrt{3}}{\sqrt{3}\cos \theta -\sin \theta }$.[3] Given that $\theta $ is a sufficiently small angle, show that $AC\approx 1+a\theta +b{{\theta }^{2}}$, for constants $a$ and $b$ to be determined exactly.[3]

Accepted Answer

(i) $AC=\frac{\sqrt3}{\sqrt3\cos\theta-\sin\theta}$. (ii) $a=\frac{\sqrt3}{3}$, $b=\frac56$.

2020 ASRJC P1 Q3

2020 ASRJC P1 Q3

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