题目

Problem

Figure 3 shows the design for a structure used to support a roof.

Figure 3

The structure consists of four wooden beams, $AB$ , $BD$ , $BC$ and $AD$ .

Given $AB=6.5$ m, $BC=BD=4.7$ m and angle $BAC=35^\circ$ ,

(a) find, to one decimal place, the size of angle $ACB$ ,

(3)

(b) find, to the nearest metre, the total length of wood required to make this structure.

(3)

解答

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在三角形 $ABC$ 中， $AB$ 对着角 $ACB$ ， $BC$ 对着角 $BAC$ 。用正弦定理，并根据图形选择钝角解。

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Using the sine rule in triangle $ABC$ ,

\begin{align*} \frac{\sin\angle ACB}{6.5} =&\,\frac{\sin35^\circ}{4.7}\\ \sin\angle ACB =&\,\frac{6.5\sin35^\circ}{4.7}\\ =&\,0.7932\ldots. \end{align*}

The acute angle is $52.5^\circ$ to 1 decimal place. From the diagram, $\angle ACB$ is obtuse, so

\begin{align*} \angle ACB =&\,180^\circ-52.5^\circ\\ =&\,127.5^\circ. \end{align*}

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总木材长度是 $AB+BC+BD+AD$ 。图中 $C$ 在 $AD$ 上，所以先求 $AC$ 和 $CD$ ，再相加得到 $AD$ 。

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In triangle $ABC$ ,

\begin{align*} \angle ABC =&\,180^\circ-35^\circ-127.5^\circ\\ =&\,17.5^\circ. \end{align*}

Using the sine rule,

\begin{align*} \frac{AC}{\sin17.5^\circ} =&\,\frac{4.7}{\sin35^\circ}\\ AC=&\,2.462\ldots. \end{align*}

In triangle $BCD$ , since $BC=BD$ , the base angles are equal:

\begin{align*} \angle BCD=\angle BDC=52.5^\circ. \end{align*}

\begin{align*} \angle CBD=180^\circ-2(52.5^\circ)=75^\circ. \end{align*}

Using the sine rule in triangle $BCD$ ,

\begin{align*} \frac{CD}{\sin75^\circ} =&\,\frac{4.7}{\sin52.5^\circ}\\ CD=&\,5.718\ldots. \end{align*}

Thus

\begin{align*} AD=AC+CD=2.462\ldots+5.718\ldots=8.180\ldots. \end{align*}

The total length of wood is

\begin{align*} 6.5+4.7+4.7+8.180\ldots =&\,24.080\ldots\\ =&\,24\text{ m} \end{align*}

to the nearest metre.