题目

Problem

The shape $ABCDA$ consists of a sector $ABCOA$ of a circle, centre $O$ , joined to a triangle $AOD$ , as shown in Figure 2.

Figure 2

The point $D$ lies on $OC$ .

The radius of the circle is $6$ cm, length $AD$ is $5$ cm and angle $AOD$ is $0.7$ radians.

(a) Find the area of the sector $ABCOA$ , giving your answer to one decimal place.

(3)

Given angle $ADO$ is obtuse,

(b) find the size of angle $ADO$ , giving your answer to 3 decimal places.

(3)

(4)

解答

(a)

解法一

思路

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要求的是大扇形 $ABCOA$ ，其圆心角是 $2\pi-0.7$ ，不是小角 $0.7$ 。

答题过程

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The angle of the sector $ABCOA$ is

\begin{align*} 2\pi-0.7. \end{align*}

Therefore the area is

\begin{align*} \frac12(6)^2(2\pi-0.7) =&\,100.497\ldots\\ =&\,100.5\text{ cm}^2. \end{align*}

(b)

解法一

思路

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在三角形 $AOD$ 中， $AD=5$ 对着角 $AOD=0.7$ ， $AO=6$ 对着角 $ADO$ 。用正弦定理。因为题目说 $\angle ADO$ 是钝角，要取钝角解。

答题过程

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Using the sine rule,

\begin{align*} \frac{\sin\angle ADO}{6} =&\,\frac{\sin0.7}{5}\\ \sin\angle ADO =&\,\frac{6\sin0.7}{5}\\ =&\,0.7730\ldots. \end{align*}

The acute angle is

\begin{align*} \sin^{-1}(0.7730\ldots)=0.884\ldots. \end{align*}

Since $\angle ADO$ is obtuse,

\begin{align*} \angle ADO =&\,\pi-0.884\ldots\\ =&\,2.258\ldots\\ =&\,2.258. \end{align*}

(c)

解法一

思路

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周长由大弧 $ABC$ 、边 $AD$ 、以及线段 $DC$ 组成。因为 $OC=6$ ，所以 $DC=OC-OD=6-OD$ 。需要先求 $OD$ 。

答题过程

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The arc length $ABC$ is

\begin{align*} 6(2\pi-0.7)=33.499\ldots. \end{align*}

In triangle $AOD$ ,

\begin{align*} \angle OAD =&\,\pi-0.7-2.258\ldots\\ =&\,0.183\ldots. \end{align*}

Using the sine rule,

\begin{align*} \frac{OD}{\sin(0.183\ldots)} =&\,\frac{5}{\sin0.7}\\ OD=&\,1.415\ldots. \end{align*}

\begin{align*} DC=6-1.415\ldots=4.584\ldots. \end{align*}

The perimeter is

\begin{align*} 33.499\ldots+5+4.584\ldots =&\,43.083\ldots\\ =&\,43.1\text{ cm}. \end{align*}

IAL 2019 May Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程