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IAL 2025 May A Q9

A Level / Edexcel / P1

IAL 2025 May A Paper · Question 9

题目

Problem

Figure 4 shows the outline of a sign that is used to advertise a bird sanctuary.

Figure 4

The sign is composed of a triangle CPQCPQ joined to a sector QCRTQQCRTQ of a circle, centre CC.

Given that

  • angle QPR=0.8QPR=0.8 radians
  • PQ=0.5 mPQ=0.5\text{ m}
  • PC=1.84 mPC=1.84\text{ m}
  • PRCPRC is a straight line

(a) find the radius, CQCQ, of the sector, in metres to 33 decimal places.

(2)

(b) Hence show that angle PCQPCQ is 0.2360.236 radians to 33 decimal places.

(2)

(c) Find the total area of the sign, giving your answer in m2\text{m}^2 to one decimal place.

(3)

(d) Find the total perimeter of the sign, giving your answer in metres to one decimal place.

(2)

解答

(a)

解法一

思路

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在三角形 PCQPCQ 中，已知 PC=1.84PC=1.84，PQ=0.5PQ=0.5，夹角 ∠QPC=0.8\angle QPC=0.8。要求 CQCQ，直接用余弦定理。

答题过程

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Using the cosine rule in triangle PCQPCQ,

CQ2= PQ2+PC2−2(PQ)(PC)cos⁡0.8= 0.52+1.842−2(0.5)(1.84)cos⁡0.8= 2.354…\begin{align*} CQ^2 =&\,PQ^2+PC^2-2(PQ)(PC)\cos0.8\\ =&\,0.5^2+1.84^2-2(0.5)(1.84)\cos0.8\\ =&\,2.354\ldots \end{align*}

So

CQ= 1.534…= 1.534 m.\begin{align*} CQ=&\,1.534\ldots\\ =&\,1.534\text{ m}. \end{align*}

(b)

解法一

思路

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接着在同一个三角形中用正弦定理。PCQPCQ 的对边是 PQ=0.5PQ=0.5，而角 QPC=0.8QPC=0.8 的对边是 CQCQ。

答题过程

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Using the sine rule in triangle PCQPCQ,

sin⁡∠PCQ0.5= sin⁡0.81.534….\begin{align*} \frac{\sin \angle PCQ}{0.5} =&\,\frac{\sin0.8}{1.534\ldots}. \end{align*}

So

sin⁡∠PCQ= 0.5sin⁡0.81.534…= 0.2337…\begin{align*} \sin \angle PCQ =&\,\frac{0.5\sin0.8}{1.534\ldots}\\ =&\,0.2337\ldots \end{align*}

Therefore

∠PCQ= 0.2359…= 0.236 radians.\begin{align*} \angle PCQ=&\,0.2359\ldots\\ =&\,0.236\text{ radians}. \end{align*}

解法二

思路

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也可以在三角形 PCQPCQ 中再次用余弦定理。这次要求夹在 PCPC 和 CQCQ 之间的角 ∠PCQ\angle PCQ。

答题过程

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Using the cosine rule,

PQ2= PC2+CQ2−2(PC)(CQ)cos⁡∠PCQ.\begin{align*} PQ^2=&\,PC^2+CQ^2-2(PC)(CQ)\cos\angle PCQ. \end{align*}

Hence

cos⁡∠PCQ= PC2+CQ2−PQ22(PC)(CQ)= 1.842+(1.534…)2−0.522(1.84)(1.534…)= 0.9723…\begin{align*} \cos\angle PCQ =&\,\frac{PC^2+CQ^2-PQ^2}{2(PC)(CQ)}\\ =&\,\frac{1.84^2+(1.534\ldots)^2-0.5^2} {2(1.84)(1.534\ldots)}\\ =&\,0.9723\ldots \end{align*}

Therefore

∠PCQ= 0.2359…= 0.236 radians.\begin{align*} \angle PCQ=&\,0.2359\ldots\\ =&\,0.236\text{ radians}. \end{align*}

(c)

解法一

思路

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标志由三角形 PCQPCQ 加上一个大扇形组成。小角 ∠PCQ=0.236\angle PCQ=0.236，所以大扇形圆心角是 2π−0.2362\pi-0.236。

总面积 = 大扇形面积 + 三角形面积。

答题过程

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The major sector has angle

2π−0.236.\begin{align*} 2\pi-0.236. \end{align*}

Using r=1.534…r=1.534\ldots, its area is

12r2θ= 12(1.534…)2(2π−0.236)= 7.114…\begin{align*} \frac12r^2\theta =&\,\frac12(1.534\ldots)^2(2\pi-0.236)\\ =&\,7.114\ldots \end{align*}

The area of triangle PCQPCQ is

12(PQ)(PC)sin⁡0.8= 12(0.5)(1.84)sin⁡0.8= 0.329…\begin{align*} \frac12(PQ)(PC)\sin0.8 =&\,\frac12(0.5)(1.84)\sin0.8\\ =&\,0.329\ldots \end{align*}

So the total area is

7.114…+0.329…= 7.444…= 7.4 m2.\begin{align*} 7.114\ldots+0.329\ldots=&\,7.444\ldots\\ =&\,7.4\text{ m}^2. \end{align*}

(d)

解法一

思路

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外周长由三段组成：大扇形弧长、PQPQ、以及直线段 CRCR。因为 P,R,CP,R,C 在同一直线上，CR=PC−CQCR=PC-CQ。

答题过程

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The major arc length is

rθ= (1.534…)(2π−0.236)= 9.276…\begin{align*} r\theta =&\,(1.534\ldots)(2\pi-0.236)\\ =&\,9.276\ldots \end{align*}

Also,

CR= PC−CQ= 1.84−1.534…= 0.305…\begin{align*} CR=&\,PC-CQ\\ =&\,1.84-1.534\ldots\\ =&\,0.305\ldots \end{align*}

Therefore the total perimeter is

9.276…+0.5+0.305…= 10.081…= 10.1 m.\begin{align*} 9.276\ldots+0.5+0.305\ldots=&\,10.081\ldots\\ =&\,10.1\text{ m}. \end{align*}