All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Using Sine Rule,

\frac{AC}{sin(\frac{3\pi}{4})} = \frac{1}{sin(\frac{\pi}{4}- \theta)}

AC = \frac{\frac{1}{\sqrt{2}}}{sin(\frac{\pi}{4})cos \theta - cos(\frac{\pi}{4})sin \theta}

AC = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}(cos \theta - sin \theta)}

AC = \frac{1}{cos \theta - sin \theta}

(ii)
For \theta is small.

AC \approx \frac{1}{1- \frac{\theta^{2}}{2} -\theta}

= [1- \frac{\theta^{2}}{2} -\theta]^{-1}

= 1 + (-1)(-\theta - \frac{\theta^{2}}{2}) + \frac{(-1)(-2)}{2!} (-\theta - \frac{\theta^{2}}{2})^{2} + \ldots

= 1 + \theta + \frac{3}{2} \theta^{2} + \ldots

KS Comments:

The first part requires students to really some O-level concepts. Students can also use the MF15 formula to evaluate it. The next part test us on our ability to do small angle approximation. At the same time, students should realise how to do the expansion of when there are three terms in the expression.

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