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P1 2026 Jan Q5

A Level / Edexcel / P1

IAL 2026 Jan Paper · Question 5

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Figure 2 shows the logo for the tourist board of an island. The logo consists of a sector BODCBBODCB of a circle centre OO joined to a triangle AODAOD.

Given that

  • ABOABO is a straight line
  • OD=3OD=3 cm
  • AD=8AD=8 cm
  • angle AOD=2.5AOD=2.5 radians

(a) show that angle ODAODA is 0.4150.415 radians to 3 significant figures.

(3)

(b) Find the total area of the logo, in cm2^2, to one decimal place.

(3)

(c) Find the perimeter of the logo, ABCDAABCDA, in cm, to one decimal place.

(3)
题目中文翻译

本题必须写出全部演算步骤。 完全依赖计算器技术的解答不予接受。

图 2 显示了某岛旅游局的标志。该标志由以 OO 为圆心的扇形 BODCBBODCB 与三角形 AODAOD 拼接而成。

已知：

  • AA、BB、OO 三点共线；
  • OD=3 cmOD=3\text{ cm}；
  • AD=8 cmAD=8\text{ cm}；
  • ∠AOD=2.5\angle AOD=2.5 弧度。

(a) 证明 ∠ODA\angle ODA 为 0.4150.415 弧度，精确至 3 位有效数字。

(b) 求该标志的总面积，单位为 cm2\text{cm}^2，精确至 1 位小数。

(c) 求该标志外周 ABCDAABCDA 的周长，单位为 cm，精确至 1 位小数。

解答

(a)

解法一

思路

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已知 ODOD 与 ADAD，并知道 ∠AOD\angle AOD，可以用正弦法则先求 ∠OAD\angle OAD，再用三角形内角和求 ∠ODA\angle ODA。这是 “show that” 题，过程要保留足够精度。

答题过程

展开

In triangle AODAOD, using the sine rule,

sin⁡(∠OAD)OD=sin⁡(∠AOD)AD\begin{align*} \frac{\sin(\angle OAD)}{OD} = \frac{\sin(\angle AOD)}{AD} \end{align*}

So

sin⁡(∠OAD)3= sin⁡2.58sin⁡(∠OAD)= 3sin⁡2.58= 0.2244…\begin{align*} \frac{\sin(\angle OAD)}{3} =&\, \frac{\sin2.5}{8}\\[3mm] \sin(\angle OAD) =&\, \frac{3\sin2.5}{8}\\[3mm] =&\, 0.2244\ldots \end{align*}

Hence

∠OAD=sin⁡−1(0.2244…)=0.2263…\begin{align*} \angle OAD = \sin^{-1}(0.2244\ldots) = 0.2263\ldots \end{align*}

Therefore

∠ODA= π−2.5−0.2263…= 0.4152…\begin{align*} \angle ODA =&\, \pi-2.5-0.2263\ldots\\[3mm] =&\, 0.4152\ldots \end{align*}

So

∠ODA=0.415\begin{align*} \angle ODA=0.415 \end{align*}

to 3 significant figures.

解法二

思路

展开

先用余弦法则求 OAOA，再用正弦法则求 ∠ODA\angle ODA。

答题过程

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Let OA=xOA=x. By the cosine rule in triangle AODAOD,

82= 32+x2−2(3)(x)cos⁡2.5x2−6xcos⁡2.5−55= 0.\begin{align*} 8^2=&\,3^2+x^2-2(3)(x)\cos2.5\\[3mm] x^2-6x\cos2.5-55=&\,0. \end{align*}

Solving this quadratic and taking the positive root gives

x=5.3924…\begin{align*} x=5.3924\ldots \end{align*}

By the sine rule,

sin⁡(∠ODA)5.3924…= sin⁡2.58sin⁡(∠ODA)= 0.4034…\begin{align*} \frac{\sin(\angle ODA)}{5.3924\ldots} =&\,\frac{\sin2.5}{8}\\[3mm] \sin(\angle ODA) =&\,0.4034\ldots \end{align*}

Hence

∠ODA= sin⁡−1(0.4034…)= 0.4152…= 0.415 radians\begin{align*} \angle ODA =&\,\sin^{-1}(0.4034\ldots)\\[3mm] =&\,0.4152\ldots\\[3mm] =&\,0.415\text{ radians} \end{align*}

to 3 significant figures.

(b)

解法一

思路

展开

总面积由扇形 BODCBBODCB 和三角形 AODAOD 组成。扇形角度不是 2.52.5，而是绕外侧的一段，所以是 2π−2.52\pi-2.5。

答题过程

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The angle of the sector is

2π−2.5\begin{align*} 2\pi-2.5 \end{align*}

The sector area is

12r2θ= 12(3)2(2π−2.5)= 17.02…\begin{align*} \frac12r^2\theta =&\, \frac12(3)^2(2\pi-2.5)\\[3mm] =&\, 17.02\ldots \end{align*}

Using ∠ODA=0.415…\angle ODA=0.415\ldots, the area of triangle AODAOD is

12(OD)(AD)sin⁡(∠ODA)= 12(3)(8)sin⁡(0.4152…)= 4.84…\begin{align*} \frac12(OD)(AD)\sin(\angle ODA) =&\, \frac12(3)(8)\sin(0.4152\ldots)\\[3mm] =&\, 4.84\ldots \end{align*}

Hence the total area is

17.02…+4.84…=21.86…\begin{align*} 17.02\ldots+4.84\ldots = 21.86\ldots \end{align*}

Therefore the total area is

21.9 cm2\begin{align*} 21.9\text{ cm}^2 \end{align*}

to one decimal place.

(c)

解法一

思路

展开

周长由线段 ABAB、外侧圆弧 BCDBCD 和线段 DADA 组成。ODOD 是扇形和三角形的公共内部边，不属于外周长。

因为 B,O,AB,O,A 共线且 OB=OD=3OB=OD=3，所以 AB=AO−3AB=AO-3。

答题过程

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The arc length BCDBCD is

rθ=3(2π−2.5)=11.35…\begin{align*} r\theta = 3(2\pi-2.5) = 11.35\ldots \end{align*}

Find AOAO using the cosine rule in triangle AODAOD:

AO2= OD2+AD2−2(OD)(AD)cos⁡(∠ODA)= 32+82−2(3)(8)cos⁡(0.4152…)= 29.07…\begin{align*} AO^2 =&\, OD^2+AD^2-2(OD)(AD)\cos(\angle ODA)\\[3mm] =&\, 3^2+8^2-2(3)(8)\cos(0.4152\ldots)\\[3mm] =&\, 29.07\ldots \end{align*}

So

AO=5.39…\begin{align*} AO=5.39\ldots \end{align*}

Since AB=AO−OBAB=AO-OB,

AB=5.39…−3=2.39…\begin{align*} AB=5.39\ldots-3=2.39\ldots \end{align*}

The perimeter is

AB+arc BCD+DA= 2.39…+11.35…+8= 21.74…\begin{align*} AB+\text{arc }BCD+DA =&\, 2.39\ldots+11.35\ldots+8\\[3mm] =&\, 21.74\ldots \end{align*}

Therefore the perimeter is

21.7 cm\begin{align*} 21.7\text{ cm} \end{align*}

to one decimal place.

解法二

思路

展开

改用正弦法则求 AOAO，再计算 AB=AO−3AB=AO-3，最后加上圆弧 BCDBCD 与 ADAD。

答题过程

展开

Using the sine rule in triangle AODAOD,

AOsin⁡(∠ODA)= ADsin⁡(∠AOD)AO= 8sin⁡(0.4152…)sin⁡2.5= 5.3924…\begin{align*} \frac{AO}{\sin(\angle ODA)} =&\,\frac{AD}{\sin(\angle AOD)}\\[3mm] AO =&\,\frac{8\sin(0.4152\ldots)}{\sin2.5}\\[3mm] =&\,5.3924\ldots \end{align*}

Therefore

AB= AO−OB= 5.3924…−3= 2.3924…\begin{align*} AB =&\,AO-OB\\[3mm] =&\,5.3924\ldots-3\\[3mm] =&\,2.3924\ldots \end{align*}

The perimeter is

AB+arc BCD+AD= 2.3924… +3(2π−2.5)+8= 21.74…\begin{align*} AB+\text{arc }BCD+AD =&\,2.3924\ldots\\[2mm] &\,+3(2\pi-2.5)+8\\[3mm] =&\,21.74\ldots \end{align*}

Hence the perimeter is

21.7 cm\begin{align*} 21.7\text{ cm} \end{align*}

to one decimal place.