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IAL 2025 May Q6

A Level / Edexcel / P1

IAL 2025 May Paper · Question 6

题目

Problem

The shaded area in Figure 2 shows the plan view of a helicopter landing pad.

Figure 2

The area consists of the major sector AOBAOB of a circle centre OO joined to a triangle AOCAOC.

Given that

  • AO=OB=15 mAO=OB=15\text{ m}
  • BC=2 mBC=2\text{ m}
  • CBOCBO is a straight line
  • angle ACO=0.6ACO=0.6 radians

(a) show that angle COACOA is 1.8471.847 radians to 33 decimal places.

(3)

(b) Find the total area of the helicopter landing pad. Give your answer in m2\text{m}^2 to 33 significant figures.

(3)

(c) Find the perimeter of the helicopter landing pad. Give your answer in metres to 33 significant figures.

(3)

解答

(a)

解法一

思路

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因为 C,B,OC,B,O 在同一直线上，OC=OB+BC=17OC=OB+BC=17。在三角形 AOCAOC 中，已知 AO=15AO=15，OC=17OC=17，以及 ∠ACO=0.6\angle ACO=0.6。先用正弦定理求 ∠CAO\angle CAO，再用三角形内角和求 ∠COA\angle COA。

答题过程

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Since CBOCBO is a straight line,

OC=OB+BC=15+2=17.\begin{align*} OC=OB+BC=15+2=17. \end{align*}

In triangle AOCAOC, using the sine rule,

sin⁡∠CAO17= sin⁡0.615.\begin{align*} \frac{\sin\angle CAO}{17} =&\,\frac{\sin0.6}{15}. \end{align*}

So

sin⁡∠CAO= 17sin⁡0.615= 0.6399…\begin{align*} \sin\angle CAO =&\,\frac{17\sin0.6}{15}\\ =&\,0.6399\ldots \end{align*}

Therefore

∠CAO=0.6944…\begin{align*} \angle CAO=0.6944\ldots \end{align*}

Now

∠COA= π−0.6−0.6944…= 1.8471…= 1.847 radians.\begin{align*} \angle COA =&\,\pi-0.6-0.6944\ldots\\ =&\,1.8471\ldots\\ =&\,1.847\text{ radians}. \end{align*}

解法二

思路

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余弦定理求边长与正弦定理求角综合法。 在斜三角形中，如果不先计算 ∠CAO\angle CAO，我们也可以先利用余弦定理求出边长 ACAC，进而利用正弦定理求出 ∠COA\angle COA：

  1. 在 △AOC\triangle AOC 中，已知边 AO=15AO = 15， OC=17OC = 17，以及对应角 ∠ACO=0.6\angle ACO = 0.6。设 AC=yAC = y。根据余弦定理： 152=y2+172−2(y)(17)cos⁡(0.6)⟹y2−34cos⁡(0.6)y+64=0\begin{align*} 15^2 = y^2 + 17^2 - 2(y)(17)\cos(0.6) \Longrightarrow y^2 - 34\cos(0.6)y + 64 = 0 \end{align*} 解此一元二次方程可得 AC≈25.56AC \approx 25.56（另一个较小的根因不合图意而舍去）；
  2. 进而使用正弦定理求 ∠COA\angle COA： sin⁡∠COAAC=sin⁡(0.6)15⟹sin⁡∠COA=ACsin⁡(0.6)15\begin{align*} \frac{\sin\angle COA}{AC} = \frac{\sin(0.6)}{15} \Longrightarrow \sin\angle COA = \frac{AC \sin(0.6)}{15} \end{align*} 由于 ∠COA\angle COA 在图中显然是钝角，取其大于 π2\frac{\pi}{2} 的解即可。这种多定理联合求解在解决复杂的斜三角形几何问题时非常通用，能拓宽学生的几何解题思路。

答题过程

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First, let AC=yAC = y. Apply the cosine rule in triangle AOCAOC to find ACAC:

AO2=  AC2+OC2−2(AC)(OC)cos⁡∠ACO152=  y2+172−2y(17)cos⁡(0.6).\begin{align*} AO^2 =&\,\, AC^2 + OC^2 - 2(AC)(OC)\cos\angle ACO\\[3mm] 15^2 =&\,\, y^2 + 17^2 - 2y(17)\cos(0.6). \end{align*}

Given cos⁡(0.6)≈0.8253\cos(0.6) \approx 0.8253, we obtain the quadratic equation:

y2−34(0.8253…)y+(289−225)=  0y2−28.061…y+64=  0.\begin{align*} y^2 - 34(0.8253\ldots)y + (289 - 225) =&\,\, 0\\[3mm] y^2 - 28.061\ldots y + 64 =&\,\, 0. \end{align*}

Solve the quadratic equation for yy:

y=  28.061…±(28.061…)2−4(64)2=  28.061…±23.053…2.\begin{align*} y =&\,\, \frac{28.061\ldots \pm \sqrt{(28.061\ldots)^2 - 4(64)}}{2}\\[3mm] =&\,\, \frac{28.061\ldots \pm 23.053\ldots}{2}. \end{align*}

Since AC>OC=17AC > OC = 17 according to the diagram, we choose the larger root:

AC=y=  25.557… m.\begin{align*} AC = y =&\,\, 25.557\ldots\text{ m}. \end{align*}

Next, apply the sine rule in triangle AOCAOC to find angle ∠COA\angle COA (let ∠COA=θ\angle COA = \theta):

sin⁡θAC=  sin⁡(0.6)AOsin⁡θ=  25.557…×sin⁡(0.6)15=  25.557…×0.5646…15=  0.9620…\begin{align*} \frac{\sin\theta}{AC} =&\,\, \frac{\sin(0.6)}{AO}\\[3mm] \sin\theta =&\,\, \frac{25.557\ldots \times \sin(0.6)}{15}\\[3mm] =&\,\, \frac{25.557\ldots \times 0.5646\ldots}{15}\\[3mm] =&\,\, 0.9620\ldots \end{align*}

Since θ\theta is an obtuse angle (θ>π2\theta > \frac{\pi}{2}), we have:

θ=  π−arcsin⁡(0.9620…)=  π−1.2944…=  1.8471… radians.\begin{align*} \theta =&\,\, \pi - \arcsin(0.9620\ldots)\\[3mm] =&\,\, \pi - 1.2944\ldots\\[3mm] =&\,\, 1.8471\ldots\text{ radians}. \end{align*}

Therefore:

∠COA=  1.847 radians\begin{align*} \angle COA =&\,\, 1.847\text{ radians} \end{align*}

to 3 decimal places.

(b)

解法一

思路

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总面积由大扇形 AOBAOB 和三角形 AOCAOC 组成。由于 ∠AOB=∠COA=1.847\angle AOB=\angle COA=1.847，大扇形的圆心角是 2π−1.8472\pi-1.847。

答题过程

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The angle of the major sector is

2π−1.847.\begin{align*} 2\pi-1.847. \end{align*}

Area of the major sector:

12r2θ= 12(15)2(2π−1.847)= 499.08…\begin{align*} \frac12r^2\theta =&\,\frac12(15)^2(2\pi-1.847)\\ =&\,499.08\ldots \end{align*}

Area of triangle AOCAOC:

12(15)(17)sin⁡1.847= 122.72…\begin{align*} \frac12(15)(17)\sin1.847 =&\,122.72\ldots \end{align*}

So the total area is

499.08…+122.72…= 621.80…= 622 m2.\begin{align*} 499.08\ldots+122.72\ldots=&\,621.80\ldots\\ =&\,622\text{ m}^2. \end{align*}

(c)

解法一

思路

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周长由大扇形弧长、线段 ACAC 和 BCBC 组成。先用余弦定理求 ACAC。

答题过程

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The major arc length is

15(2π−1.847)=66.54…\begin{align*} 15(2\pi-1.847)=66.54\ldots \end{align*}

Using the cosine rule in triangle AOCAOC,

AC2= 152+172−2(15)(17)cos⁡1.847= 653.13…\begin{align*} AC^2 =&\,15^2+17^2-2(15)(17)\cos1.847\\ =&\,653.13\ldots \end{align*}

So

AC=25.55…\begin{align*} AC=25.55\ldots \end{align*}

Therefore the perimeter is

66.54…+25.55…+2= 94.09…= 94.1 m.\begin{align*} 66.54\ldots+25.55\ldots+2=&\,94.09\ldots\\ =&\,94.1\text{ m}. \end{align*}