The diagram shows a quadrilateral field $ABCD$, where $AB = 21$ m and $BC = 3$ m. Angle \

Question

The diagram shows a quadrilateral field $ABCD$, where $AB = 21$ m and $BC = 3$ m. Angle $\angle ABC = \angle ADC = 90{}^\circ$. Angle $\angle BAD = \theta$, for $0{}^\circ < \theta < 90{}^\circ$, and can vary. The perimeter of the fencing around the quadrilateral field $ABCD$ is $P$ m. Show that $P = 24 + 18 \cos \theta + 24 \sin \theta$.[3] Express $P$ in the form $R \sin(\theta + \alpha) + 24$, where $R > 0$ and $0{}^\circ < \alpha < 90{}^\circ$.[3] Given that the total perimeter of the fencing is 53 m, find the value(s) of $\theta$.[2] Explain why the total length of the fencing will never exceed a certain value and state this value.[2]

Accepted Answer

(i) $P=24+18\cos\theta+24\sin\theta$ shown (ii) $P=30\sin(\theta+36.9^\circ)+24$ (iii) $\theta=38.3^\circ$ or $68.0^\circ$ (iv) $54\text{ m}$

(A Math) 2025 CCHY PRELIMS P1 Q10

(A Math) 2025 CCHY PRELIMS P1 Q10

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