 Problem of the Week

Problem E and Solution

Problem

Tima owns a triangular parcel of land that is created by three intersecting roads, as shown. Two of the roads meet at a right angle and two of the roads intersect at a $$25^{\circ}$$ angle. If the perimeter of the triangular parcel of land is 1000 m, what is its area to the nearest 100 m$$^2$$? Recall: For any acute angle, $$\theta$$, of any right-angled triangle, we define the following:

sin $$\theta= \frac{\mbox{opp}}{\mbox{hyp}}$$, cos $$\theta= \frac{\mbox{adj}}{\mbox{hyp}}$$, tan $$\theta= \frac{\mbox{opp}}{\mbox{adj}}$$

Solution

We will label the diagram as shown. We know the following:
$$a+b+c=1000$$ (1)
$$\dfrac{c}{b} = \sin(25^{\circ})$$, and so $$c=b \sin(25^{\circ})$$ (2)
$$\dfrac{a}{b} = \cos(25^{\circ})$$, and so $$a=b \cos(25^{\circ})$$ (3)
Substituting (2) and (3) into (1) we get: \begin{aligned} b\cos(25^{\circ}) + b + b\sin(25^{\circ}) &=& 1000\\ b(\cos(25^{\circ}) + 1 + \sin(25^{\circ})) &=& 1000\\ b &=& \frac{1000}{\cos(25^{\circ}) + 1 + \sin(25^{\circ}) }\end{aligned}

$$b \hspace{2mm} \approx \hspace{2mm} 429.38$$ m
Now, since $$a=b\sin(25^{\circ})$$ and $$c=b\cos(25^{\circ})$$, the area of the triangle is
\begin{aligned} %\dfrac{ac}{2} \approx \dfrac{(181.5)(389.2)}{2} = 35\,319.9\\[1mm] \dfrac{ac}{2} &=& \dfrac{(b\sin(25^{\circ}))(b\cos(25^{\circ}))}{2}\\ & =& \dfrac{b^2\sin(25^{\circ})\cos(25^{\circ})}{2}\\ &\approx& \dfrac{(429.38)^2\sin(25^{\circ})\cos(25^{\circ})}{2} \\ &=& 35\,308.4 \mbox{ m^2}\end{aligned} Therefore, to the nearest 100 m$$^2$$, the area of the triangle is 35 300 m$$^2$$.