题目

Problem

Figure 2 shows the plan view of a frame for a flat roof.

Figure 2

The shape of the frame consists of triangle $ABD$ joined to triangle $BCD$ .

Given that

$BD=x$ m
$CD=(1+x)$ m
$BC=5$ m
angle $BCD=\theta^\circ$

(a) show that

\begin{align*} \cos\theta^\circ=\frac{13+x}{5+5x}. \end{align*}

(2)

Given also that

$x=2\sqrt3$
angle $BAC=30^\circ$
$ADC$ is a straight line

(b) find the area of triangle $ABC$ , giving your answer, in m $^2$ , to one decimal place.

(5)

解答

(a)

解法一

思路

展开

在三角形 $BCD$ 中，已知三边 $BD=x$ 、 $CD=1+x$ 、 $BC=5$ ，要求夹角 $\theta$ 的余弦，直接用余弦定理。

答题过程

展开

Using the cosine rule in triangle $BCD$ ,

\begin{align*} BD^2=&\,BC^2+CD^2-2(BC)(CD)\cos\theta^\circ\\ x^2=&\,5^2+(1+x)^2-2(5)(1+x)\cos\theta^\circ. \end{align*}

Rearrange:

\begin{align*} 2(5)(1+x)\cos\theta^\circ =&\,25+(1+x)^2-x^2\\ =&\,25+1+2x+x^2-x^2\\ =&\,26+2x. \end{align*}

Therefore

\begin{align*} \cos\theta^\circ =&\,\frac{26+2x}{10(1+x)}\\ =&\,\frac{2(13+x)}{10(1+x)}\\ =&\,\frac{13+x}{5+5x}. \end{align*}

(b)

解法一

思路

展开

先代入 $x=2\sqrt3$ 求出 $\theta$ 。因为 $A,D,C$ 共线，所以 $\angle ACB=\theta$ 。在三角形 $ABC$ 中，已知 $\angle A=30^\circ$ 、 $\angle C=\theta$ 和边 $BC=5$ ，可用正弦定理求 $AB$ ，再用 $\frac12 ab\sin C$ 求面积。

答题过程

展开

Substitute $x=2\sqrt3$ into the result from part (a):

\begin{align*} \cos\theta^\circ =&\,\frac{13+2\sqrt3}{5+10\sqrt3}. \end{align*}

Hence

\begin{align*} \theta=42.4707\ldots^\circ. \end{align*}

Since $A,D,C$ is a straight line,

\begin{align*} \angle ACB=\theta. \end{align*}

In triangle $ABC$ ,

\begin{align*} \angle ABC =&\,180^\circ-30^\circ-\theta\\ =&\,107.5292\ldots^\circ. \end{align*}

Using the sine rule,

\begin{align*} \frac{AB}{\sin\theta} =&\,\frac{BC}{\sin30^\circ}\\ AB =&\,\frac{5\sin\theta}{\sin30^\circ}\\ =&\,6.7519\ldots. \end{align*}

Therefore

\begin{align*} \text{Area of }ABC =&\,\frac12(AB)(BC)\sin\angle ABC\\ =&\,\frac12(6.7519\ldots)(5)\sin(107.5292\ldots^\circ)\\ =&\,16.101\ldots\\ =&\,16.1\text{ m}^2. \end{align*}

IAL 2023 Oct Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程