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IAL 2025 Jan Q8

A Level / Edexcel / P1

IAL 2025 Jan Paper · Question 8

题目

Problem

Figure 1 shows a sketch of a design for a badge.

Figure 1

The design consists of a triangle OABOAB joined to a sector OBCOBC of a circle with centre OO.

In the design

  • OB=3.4 cmOB=3.4\text{ cm}
  • AB=1.9 cmAB=1.9\text{ cm}
  • angle AOB=π6AOB=\dfrac{\pi}{6} radians
  • angle OAB>π2OAB>\dfrac{\pi}{2} radians

Making your method clear,

(a) find the size of angle OABOAB, giving your answer in radians to 44 significant figures.

(3)

(b) find the area of triangle OABOAB, in cm2\text{cm}^2, giving your answer to 33 significant figures.

(2)

Given that the ratio of the area of sector OBCOBC to the area of triangle OABOAB is 3:23:2,

(c) show that angle BOCBOC is 0.4620.462 radians to 33 significant figures.

(3)

(d) Hence find the perimeter of the badge, in cm, to the nearest integer.

(5)

解答

(a)

解法一

思路

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在三角形 OABOAB 中，ABAB 对应角 AOB=π6AOB=\frac{\pi}{6}，OBOB 对应角 OABOAB。用正弦定理可以先求 sin⁡∠OAB\sin\angle OAB。由于题目给出 ∠OAB>π2\angle OAB>\frac{\pi}{2}，所以要取钝角解。

答题过程

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Using the sine rule in triangle OABOAB,

sin⁡∠OAB3.4= sin⁡(π/6)1.9.\begin{align*} \frac{\sin\angle OAB}{3.4} =&\,\frac{\sin(\pi/6)}{1.9}. \end{align*}

So

sin⁡∠OAB= 3.4sin⁡(π/6)1.9= 0.8947…\begin{align*} \sin\angle OAB =&\,\frac{3.4\sin(\pi/6)}{1.9}\\ =&\,0.8947\ldots \end{align*}

The principal angle is

sin⁡−1(0.8947…)=1.1078…\begin{align*} \sin^{-1}(0.8947\ldots)=1.1078\ldots \end{align*}

But ∠OAB>π2\angle OAB>\frac{\pi}{2}, so

∠OAB= π−1.1078…= 2.0337…= 2.034 radians.\begin{align*} \angle OAB =&\,\pi-1.1078\ldots\\ =&\,2.0337\ldots\\ =&\,2.034\text{ radians}. \end{align*}

(b)

解法一

思路

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先求第三个角 ∠OBA\angle OBA，再用面积公式 12absin⁡C\frac12ab\sin C。这里夹角可以用

π−π6−∠OAB.\begin{align*} \pi-\frac{\pi}{6}-\angle OAB. \end{align*}

答题过程

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The angle between OBOB and ABAB is

∠OBA= π−π6−2.0337…= 0.5842…\begin{align*} \angle OBA =&\,\pi-\frac{\pi}{6}-2.0337\ldots\\ =&\,0.5842\ldots \end{align*}

Therefore

Area of triangle OAB= 12(3.4)(1.9)sin⁡(0.5842…)= 1.7815…= 1.78 cm2.\begin{align*} \text{Area of triangle }OAB =&\,\frac12(3.4)(1.9)\sin(0.5842\ldots)\\ =&\,1.7815\ldots\\ =&\,1.78\text{ cm}^2. \end{align*}

(c)

解法一

思路

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扇形面积 : 三角形面积 = 3:23:2，所以扇形面积是三角形面积的 32\frac32。再用扇形面积公式 12r2θ\frac12r^2\theta 求 θ=∠BOC\theta=\angle BOC。

答题过程

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The area of sector OBCOBC is

32(1.7815…)=2.6723…\begin{align*} \frac32(1.7815\ldots)=2.6723\ldots \end{align*}

Using the sector area formula,

12(3.4)2θ= 2.6723…θ= 2(2.6723…)(3.4)2= 0.4623…\begin{align*} \frac12(3.4)^2\theta=&\,2.6723\ldots\\ \theta=&\,\frac{2(2.6723\ldots)}{(3.4)^2}\\ =&\,0.4623\ldots \end{align*}

So

∠BOC=0.462 radians.\begin{align*} \angle BOC=0.462\text{ radians}. \end{align*}

(d)

解法一

思路

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周长由 OAOA、ABAB、弧 BCBC、以及 OCOC 组成。因为扇形半径是 3.43.4，所以 OC=3.4OC=3.4，弧长 BC=3.4θBC=3.4\theta。还需要用余弦定理求 OAOA。

答题过程

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First find OAOA using the cosine rule:

OA2= 3.42+1.92−2(3.4)(1.9)cos⁡(0.5842…)= 4.3930…\begin{align*} OA^2 =&\,3.4^2+1.9^2 -2(3.4)(1.9)\cos(0.5842\ldots)\\ =&\,4.3930\ldots \end{align*}

So

OA=2.0959…\begin{align*} OA=2.0959\ldots \end{align*}

The arc length BCBC is

3.4(0.4623…)=1.5719…\begin{align*} 3.4(0.4623\ldots)=1.5719\ldots \end{align*}

Therefore the perimeter is

OA+AB+arc BC+OC= 2.0959…+1.9+1.5719…+3.4= 8.9678…= 9 cm.\begin{align*} OA+AB+\text{arc }BC+OC =&\,2.0959\ldots+1.9+1.5719\ldots+3.4\\ =&\,8.9678\ldots\\ =&\,9\text{ cm}. \end{align*}

解法二

思路

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正弦定理求边长法。 在第一问 (a) 中，我们已经通过正弦定理求出了 ∠OAB=2.0337…\angle OAB = 2.0337\ldots 弧度，进而求出第三个内角 ∠OBA=0.5842…\angle OBA = 0.5842\ldots 弧度。 当三角形中所有内角均已知，且有已知边长（如 AB=1.9AB = 1.9）时，求第三边 OAOA 的最快方法是再次使用正弦定理：

OAsin⁡∠OBA=ABsin⁡∠AOB⟹OA=1.9sin⁡(0.5842…)sin⁡(π/6)\begin{align*} \frac{OA}{\sin\angle OBA} = \frac{AB}{\sin\angle AOB} \Longrightarrow OA = \frac{1.9 \sin(0.5842\ldots)}{\sin(\pi/6)} \end{align*}

这比起余弦定理中含有平方和乘积的多项式计算要精简十倍，而且仅涉及一步乘除法运算，极大地降低了计算出错的概率，是考场上非常高效的拿分技巧。

答题过程

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Apply the sine rule in triangle OABOAB to find the length of OAOA:

OAsin⁡∠OBA=  ABsin⁡∠AOB.\begin{align*} \frac{OA}{\sin\angle OBA} =&\,\, \frac{AB}{\sin\angle AOB}. \end{align*}

Using AB=1.9AB = 1.9, ∠AOB=π6\angle AOB = \frac{\pi}{6}, and ∠OBA=0.5842…\angle OBA = 0.5842\ldots from part (b):

OA=  1.9×sin⁡(0.5842…)sin⁡(π/6)=  1.9×0.5515…0.5=  3.8×0.5515…=  2.0959… cm.\begin{align*} OA =&\,\, \frac{1.9 \times \sin(0.5842\ldots)}{\sin(\pi/6)}\\[3mm] =&\,\, \frac{1.9 \times 0.5515\ldots}{0.5}\\[3mm] =&\,\, 3.8 \times 0.5515\ldots\\[3mm] =&\,\, 2.0959\ldots\text{ cm}. \end{align*}

The arc length BCBC is:

arc BC=  3.4×0.4623…=  1.5719… cm.\begin{align*} \text{arc }BC =&\,\, 3.4 \times 0.4623\ldots\\[3mm] =&\,\, 1.5719\ldots\text{ cm}. \end{align*}

Therefore, the perimeter of the badge is:

Perimeter=  OA+AB+arc BC+OC=  2.0959…+1.9+1.5719…+3.4=  8.9678…≈  9 cm\begin{align*} \text{Perimeter} =&\,\, OA + AB + \text{arc }BC + OC\\[3mm] =&\,\, 2.0959\ldots + 1.9 + 1.5719\ldots + 3.4\\[3mm] =&\,\, 8.9678\ldots\\[3mm] \approx&\,\, 9\text{ cm} \end{align*}

to the nearest integer.