In the triangle $ABC$, angle $BAC=\theta \mathrm{ radians,}$ angle \(ACB=\frac{\pi }{6}\mathrm{

Question

In the triangle $ABC$, angle $BAC=\theta \mathrm{ radians,}$ angle $ACB=\frac{\pi }{6}\mathrm{ radians}$ and $AC=\sqrt{3}.$ Given that $\theta $ is sufficiently small, show that $AB\approx \frac{2\sqrt{3}}{2+2\sqrt{3}\theta -{{\theta }^{2}}}\approx a+b\theta +c{{\theta }^{2}},$ where $a$, $b$ and $c$ are constants to be determined in exact form.[7]

Accepted Answer

a=sqrt(3), b=-3, c=7sqrt(3)/2.

2015 AJC P1 Q5

2015 AJC P1 Q5

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