2025 MSHS PRELIMS P2 Q10
2025 MSHS PRELIMS P2 Q10
Secondary 4
11 marks
A L-shaped structure, \(OPQ\) can be rotated about \(O\). \(OP\) and \(PQ\) measures 5 m and 1 m respectively. \(OP\) makes an acute angle, \(\theta\), with the ground. The shortest distance from \(Q\) to the wall is \(L\) m.
- Show that \(L = 5\cos \theta - \sin \theta\).[2]
- Express \(L\) in the form \(R \cos (\theta + \alpha)\), where \(R>0\) and \(0^\circ < \alpha < 90^\circ\).[3]
- State the minimum value of \(L\) and the corresponding value of \(\theta\).[3]
- Find the value of \(\theta\) when \(L = 3\).[2]
- Explain why the maximum value of \(L\) is not \(R\).[1]
Solution:
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Answer:(a) \(L=5\cos\theta-\sin\theta\) (b) \(L=\sqrt{26}\cos\left(\theta+\tan^{-1}\left(\frac{1}{5}\right)\right)\) (c) \(0\text{ m}\), when \(\theta=78.7^\circ\) (d) \(\theta=42.7^\circ\) (e) \(L=R\) would require \(\theta+\alpha=0^\circ\), but \(\theta+\alpha>0^\circ\).
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