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2026 NYGH P2 Q13

2026 NYGH P2 Q13

IP 4
11 marks

The diagram below shows a figure \(ABCDE\) where \(\angle ACB\) is \(90^\circ\). \(BCD\) is a straight line. Given that \(AB=16\text{ cm}\), \(DE=6\text{ cm}\), \(\angle BAC=\theta\) and \(\angle CDE=\theta+30^\circ\) where \(0^\circ<\theta<60^\circ\).

  1. Show that \(AE=\left(13\cos\theta-3\sqrt3\sin\theta\right)\text{ cm}\).[2]
  2. The perimeter of the figure \(ABCDE\) is \(P\text{ cm}\). Show that \(P=\left(13-3\sqrt3\right)\sin\theta+\left(13+3\sqrt3\right)\cos\theta+22\).[4]
  3. Express \(P\) in the form of \(R\sin(\theta+\alpha)+22\) where \(R>0\) and \(\alpha\) is an acute angle.[3]
  4. Find the maximum value of \(P\) and the corresponding value of \(\theta\).[2]

Solution:

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Answer:(a) \(AE=(13\cos\theta-3\sqrt3\sin\theta)\text{ cm}\) (b) \(P=(13-3\sqrt3)\sin\theta+(13+3\sqrt3)\cos\theta+22\) (c) \(P=14\sqrt2\sin(\theta+66.8^\circ)+22\) (d) \(P_{\max}=22+14\sqrt2\approx41.8\), \(\theta=23.2^\circ\)

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