2016 VJC Promo Q2

2016 VJC Promo Q2

6 marks
Promo

The diagram below shows a circle with centre \(C\) and radius \(4\) cm. The line \(AB\) is a tangent to the circle and angle \(ACB=\left( \frac{\pi }{3}-\theta \right)\) radians.

Show that \(A\), the area of the shaded region \(R\), is given by \(8\left[ \tan \left( \frac{\pi }{3}-\theta \right)-\left( \frac{\pi }{3}-\theta \right) \right]\).[2]

Given that \(\theta \) is a sufficiently small angle, show that \(A\approx \left( 8\sqrt{3}-\frac{8\pi }{3} \right)+a\theta +b{{\theta }^{2}}\), where \(a\) and \(b\) are constants to be determined.[4]

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Answer:\(A=8\left[\tan\left(\frac{\pi}{3}-\theta\right)-\left(\frac{\pi}{3}-\theta\right)\right]\), and \(A\approx\left(8\sqrt3-\frac{8\pi}{3}\right)-24\theta+32\sqrt3\theta^2\). Hence \(a=-24\) and \(b=32\sqrt3\).

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