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P1 2021 Oct Q7

A Level / Edexcel / P1

IAL 2021 Oct Paper · Question 7

题目

Problem

Figure 3 shows the design for a sign at a bird sanctuary.

Figure 3

The design consists of a kite OABCOABC joined to a sector OCXAOCXA of a circle centre OO.

In the design

  • OA=OC=0.6OA=OC=0.6 m
  • AB=CB=1.4AB=CB=1.4 m
  • Angle OAB=OAB= Angle OCB=2OCB=2 radians
  • Angle AOC=θAOC=\theta radians, as shown in Figure 3

Making your method clear,

(a) show that θ=1.64\theta=1.64 radians to 3 significant figures,

(4)

(b) find the perimeter of the sign, in metres to 2 significant figures,

(2)

(c) find the area of the sign, in m2\text{m}^2 to 2 significant figures.

(4)

解答

(a)

解法一

思路

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先在三角形 OABOAB 中用余弦定理求 OBOB，再用正弦定理求角 AOBAOB。由于图形是对称的，θ=2∠AOB\theta=2\angle AOB。

答题过程

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In triangle OABOAB, use the cosine rule:

OB2= 0.62+1.42−2(0.6)(1.4)cos⁡2= 3.019…,\begin{align*} OB^2 =&\,0.6^2+1.4^2-2(0.6)(1.4)\cos2\\ =&\,3.019\ldots, \end{align*}

so

OB=1.737….\begin{align*} OB=1.737\ldots. \end{align*}

Using the sine rule in triangle OABOAB,

sin⁡∠AOB1.4= sin⁡21.737…sin⁡∠AOB= 1.4sin⁡21.737…∠AOB= 0.822….\begin{align*} \frac{\sin\angle AOB}{1.4} =&\,\frac{\sin2}{1.737\ldots}\\ \sin\angle AOB =&\,\frac{1.4\sin2}{1.737\ldots}\\ \angle AOB=&\,0.822\ldots. \end{align*}

By symmetry,

θ= 2∠AOB= 2(0.822…)= 1.644…= 1.64to 3 significant figures.\begin{align*} \theta =&\,2\angle AOB\\ =&\,2(0.822\ldots)\\ =&\,1.644\ldots\\ =&\,1.64\quad\text{to 3 significant figures.} \end{align*}

(b)

解法一

思路

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外边界由两条长边 AB,CBAB,CB 和大弧 CXACXA 组成。大弧的圆心角是 2π−θ2\pi-\theta。

答题过程

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The major arc CXACXA has angle

2π−1.64.\begin{align*} 2\pi-1.64. \end{align*}

Its length is

0.6(2π−1.64).\begin{align*} 0.6(2\pi-1.64). \end{align*}

Therefore the perimeter is

1.4+1.4+0.6(2π−1.64)= 5.585…= 5.6.\begin{align*} 1.4+1.4+0.6(2\pi-1.64) =&\,5.585\ldots\\ =&\,5.6. \end{align*}

So the perimeter is

5.6 m.\begin{align*} 5.6\text{ m}. \end{align*}

(c)

解法一

思路

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总面积由大扇形 OCXAOCXA 加上风筝 OABCOABC。风筝可分成两个全等三角形，每个三角形有边 0.60.6 和 1.41.4，夹角为 22。

答题过程

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Area of the major sector OCXAOCXA:

12(0.6)2(2π−1.64).\begin{align*} \frac12(0.6)^2(2\pi-1.64). \end{align*}

Area of the kite OABCOABC:

2(12(0.6)(1.4)sin⁡2)=0.6(1.4)sin⁡2.\begin{align*} 2\left(\frac12(0.6)(1.4)\sin2\right) =0.6(1.4)\sin2. \end{align*}

Therefore the total area is

12(0.6)2(2π−1.64)+0.6(1.4)sin⁡2= 1.599…= 1.6.\begin{align*} \frac12(0.6)^2(2\pi-1.64)+0.6(1.4)\sin2 =&\,1.599\ldots\\ =&\,1.6. \end{align*}

So the area is

1.6 m2.\begin{align*} 1.6\text{ m}^2. \end{align*}