In the triangle $ABC$, $AC=\sqrt{3}$, angle $BAC=\frac{2\pi }{3}$ radians and $ACB=\theta $

Question

In the triangle $ABC$, $AC=\sqrt{3}$, angle $BAC=\frac{2\pi }{3}$ radians and $ACB=\theta $ radians (see diagram). Show that $BC=\frac{3}{\sqrt{3}\cos \theta -\sin \theta }$ .[3] Given that $\theta $ is a sufficiently small angle, show that $BC\approx \sqrt{3}+\theta +a{{\theta }^{2}}$, where $a$ is a constant to be determined.[4]

Accepted Answer

$BC = \sqrt{3} + \theta + \dfrac{5\sqrt{3}}{6}\theta^2 + \dots$

2017 MI Promo Q11

2017 MI Promo Q11

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