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IAL 2021 Jan Q5

A Level / Edexcel / P1

IAL 2021 Jan Paper · Question 5

题目

Problem

Figure 3 shows the plan view of a viewing platform at a tourist site.

Figure 3

The shape of the viewing platform consists of a sector ABCOAABCOA of a circle, centre OO, joined to a triangle AODAOD.

Given that

  • OA=OC=6OA=OC=6 m
  • AD=14AD=14 m
  • angle ADC=0.43ADC=0.43 radians
  • angle AODAOD is an obtuse angle
  • OCDOCD is a straight line

find

(a) the size of angle AODAOD, in radians, to 3 decimal places,

(3)

(b) the length of arc ABCABC, in metres, to one decimal place,

(2)

(c) the total area of the viewing platform, in m2\text{m}^2 to one decimal place.

(4)

解答

(a)

解法一

思路

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因为 O,C,DO,C,D 共线，三角形 AODAOD 中已知 OA=6OA=6、AD=14AD=14，以及角 ADC=0.43ADC=0.43。角 ADOADO 也是 0.430.43。先用正弦定理求角 AODAOD 的补角，再用题目给出的 obtuse 条件选择钝角。

答题过程

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In triangle AODAOD,

∠ADO=0.43.\begin{align*} \angle ADO=0.43. \end{align*}

Using the sine rule,

sin⁡α14= sin⁡0.436sin⁡α= 14sin⁡0.436α= 1.337….\begin{align*} \frac{\sin\alpha}{14} =&\,\frac{\sin0.43}{6}\\ \sin\alpha =&\,\frac{14\sin0.43}{6}\\ \alpha=&\,1.337\ldots. \end{align*}

Since ∠AOD\angle AOD is obtuse,

∠AOD= π−1.337…= 1.805…= 1.805.\begin{align*} \angle AOD =&\,\pi-1.337\ldots\\ =&\,1.805\ldots\\ =&\,1.805. \end{align*}

(b)

解法一

思路

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扇形 ABCOAABCOA 是大扇形，圆心角是 2π−∠AOD2\pi-\angle AOD。弧长公式是 s=rθs=r\theta。

答题过程

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The angle for the major arc ABCABC is

2π−1.805….\begin{align*} 2\pi-1.805\ldots. \end{align*}

Thus

arc ABC= 6(2π−1.805…)= 26.86…= 26.9 m.\begin{align*} \text{arc }ABC =&\,6(2\pi-1.805\ldots)\\ =&\,26.86\ldots\\ =&\,26.9\text{ m}. \end{align*}

(c)

解法一

思路

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总面积由大扇形 ABCOAABCOA 和三角形 AODAOD 组成。扇形用 12r2θ\frac12r^2\theta，三角形用 12absin⁡C\frac12ab\sin C。三角形中夹角是 ∠OAD=π−0.43−∠AOD\angle OAD=\pi-0.43-\angle AOD。

答题过程

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The sector angle is

2π−1.805….\begin{align*} 2\pi-1.805\ldots. \end{align*}

Area of the sector:

12(6)2(2π−1.805…)=80.60….\begin{align*} \frac12(6)^2(2\pi-1.805\ldots)=80.60\ldots. \end{align*}

In triangle AODAOD,

∠OAD= π−0.43−1.805…= 0.906….\begin{align*} \angle OAD =&\,\pi-0.43-1.805\ldots\\ =&\,0.906\ldots. \end{align*}

Area of triangle AODAOD:

12(6)(14)sin⁡(0.906…)=33.09….\begin{align*} \frac12(6)(14)\sin(0.906\ldots)=33.09\ldots. \end{align*}

Therefore the total area is

80.60…+33.09…= 113.69…= 113.7 m2.\begin{align*} 80.60\ldots+33.09\ldots =&\,113.69\ldots\\ =&\,113.7\text{ m}^2. \end{align*}