In triangle $PQR$, $\angle PRQ=\frac{2\pi }{3}$, $\angle RPQ=\frac{\pi }{6}+\theta $, and \(PR

Question

In triangle $PQR$, $\angle PRQ=\frac{2\pi }{3}$, $\angle RPQ=\frac{\pi }{6}+\theta $, and $PR=a$ units, where $a$ is a positive real constant. Show that $PQ=\frac{\sqrt{3}a}{\cos \theta -\sqrt{3}\sin \theta }$.[4] Given that $\theta $ is sufficiently small, show that $PQ\approx \sqrt{3}a\left( 1+b\theta +c{{\theta }^{2}} \right)$, where $b$ and $c$ are real constants.[3] Given that $\theta >0$, find the range of values of $\theta $ such that the percentage error of the approximation in (ii) is less than $5\%$.[2]

Accepted Answer

(ii) $PQ\approx \sqrt{3}a\left( 1+\sqrt{3}	heta +\frac{7}{2}{{	heta }^{2}} ight)$ (iii) $\mathrm{f}\left( 	heta ight)<0.05$

2023 SAJC P2 Q2

2023 SAJC P2 Q2

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