A gardener is trying to fence a garden. The interior of the quadrilateral $ABCD$ represents her ga

Question

A gardener is trying to fence a garden. The interior of the quadrilateral $ABCD$ represents her garden where $AB$ and $BC$ are walls that are perpendicular to each other. The two fences are such that $AD = 8$ m, $DC = 6$ m and $\angle BAD = \angle DCB = \theta$, where $\theta$ is an acute angle that varies. Show that the area of the garden, $A$ m$^2$, is given by $A = 25 \sin 2\theta - 24 \cos 2\theta + 24$. Express $A$ in the form of $R \sin (2\theta - \alpha) + 24$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. If the gardener wants to fence up an area of $40$ m$^2$, find the value of $\theta$. With the above dimensions, can the gardener mark out an area of $60$ m$^2$? Justify your answer with clear workings.

Accepted Answer

(i) $A=25\sin2\theta-24\cos2\theta+24$ (shown). (ii) $A=\sqrt{1201}\sin(2\theta-\alpha)+24$, $\alpha=\tan^{-1}\!\left(\dfrac{24}{25}\right)\approx43.8^\circ$. (iii) $\theta\approx35.7^\circ$, a concave configuration. (iv) No: $A_{\max}=24+\sqrt{1201}\approx58.7\text{ m}^2<60\text{ m}^2$.

Trigonometry - Trigonometric Modelling

Trigonometry - Trigonometric Modelling

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