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IAL 2026 Jan A Q6

A Level / Edexcel / P1

IAL 2026 Jan A Paper · Question 6

题目

Problem

In this question you must show all stages of your working.

Figure 3 shows the design for a shop sign ABCDAABCDA.

Figure 3

The sign consists of a triangle AODAOD joined to a sector of a circle DOBCDDOBCD with radius 1.81.8 m and centre OO.

The points AA, BB and OO lie on a straight line.

Given that AD=3.9AD=3.9 m and angle BODBOD is 0.840.84 radians,

(a) calculate the size of angle DAODAO, giving the answer in radians to 3 decimal places.

(2)

(b) Find, in m, the length of AOAO giving the answer to 2 decimal places.

(3)

(c) Find, in m2^2, the area of the shop sign, giving the answer to one decimal place.

(3)

(d) Find, in m, the perimeter of the shop sign, giving the answer to one decimal place.

(3)

解答

(a)

解法一

思路

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因为 A,B,OA,B,O 共线，所以 ∠AOD=0.84\angle AOD=0.84。在三角形 AODAOD 中，已知 ODOD、ADAD 和 ∠AOD\angle AOD，可以用正弦法则求 ∠DAO\angle DAO。

答题过程

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Since AA, BB and OO lie on a straight line,

∠AOD=0.84.\begin{align*} \angle AOD=0.84. \end{align*}

Using the sine rule in triangle AODAOD,

sin⁡∠DAO1.8= sin⁡0.843.9sin⁡∠DAO= 1.8sin⁡0.843.9∠DAO= 0.3507…\begin{align*} \frac{\sin\angle DAO}{1.8} =&\,\frac{\sin0.84}{3.9}\\ \sin\angle DAO =&\,\frac{1.8\sin0.84}{3.9}\\ \angle DAO =&\,0.3507\ldots \end{align*}

Therefore

∠DAO=0.351 radians\begin{align*} \angle DAO=0.351\text{ radians} \end{align*}

to 3 decimal places.

(b)

解法一

思路

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先由三角形内角和求 ∠ADO\angle ADO，再对 AOAO 使用余弦法则。这里要保留 (a) 的未四舍五入值参与计算，最后才把 AOAO 四舍五入到 2 位小数。

答题过程

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Using

∠DAO=0.3507…,\begin{align*} \angle DAO=0.3507\ldots, \end{align*}

we have

∠ADO= π−0.84−0.3507…= 1.9508…\begin{align*} \angle ADO =&\,\pi-0.84-0.3507\ldots\\ =&\,1.9508\ldots \end{align*}

By the cosine rule,

AO2= 1.82+3.92 −2(1.8)(3.9)cos⁡(1.9508…)AO= 4.860…\begin{align*} AO^2 =&\,1.8^2+3.9^2\\ &\,\hspace{2pt}-2(1.8)(3.9)\cos(1.9508\ldots)\\ AO=&\,4.860\ldots \end{align*}

Therefore

AO=4.86 m\begin{align*} AO=4.86\text{ m} \end{align*}

to 2 decimal places.

解法二

思路

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也可以把 AOAO 看成两个水平投影的和。这个方法在图形题里很直观：从 DD 向 AOAO 作垂线，AOAO 就由 ODOD 的投影和 ADAD 的投影相加得到。

答题过程

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Resolve the two sides along the direction of AOAO.

The horizontal component of ODOD is

1.8cos⁡0.84.\begin{align*} 1.8\cos0.84. \end{align*}

The horizontal component of ADAD is

3.9cos⁡(0.3507…).\begin{align*} 3.9\cos(0.3507\ldots). \end{align*}

Hence

AO= 1.8cos⁡0.84+3.9cos⁡(0.3507…)= 4.860…\begin{align*} AO =&\,1.8\cos0.84+3.9\cos(0.3507\ldots)\\ =&\,4.860\ldots \end{align*}

Therefore

AO=4.86 m.\begin{align*} AO=4.86\text{ m}. \end{align*}

(c)

解法一

思路

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整个招牌面积等于大扇形 DOBCDDOBCD 的面积加上三角形 AODAOD 的面积。扇形用角度 2π−0.842\pi-0.84，三角形可以用 12absin⁡C\frac12ab\sin C。

答题过程

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The angle of the major sector is

2π−0.84.\begin{align*} 2\pi-0.84. \end{align*}

So the area of the sector is

12(1.8)2(2π−0.84).\begin{align*} \frac12(1.8)^2(2\pi-0.84). \end{align*}

The area of triangle AODAOD is

12(1.8)(AO)sin⁡0.84.\begin{align*} \frac12(1.8)(AO)\sin0.84. \end{align*}

Using AO=4.860…AO=4.860\ldots,

area= 12(1.8)2(2π−0.84) +12(1.8)(4.860…)sin⁡0.84= 12.08…\begin{align*} \text{area} =&\,\frac12(1.8)^2(2\pi-0.84)\\ &\,\hspace{2pt}+\frac12(1.8)(4.860\ldots)\sin0.84\\ =&\,12.08\ldots \end{align*}

Therefore the area is

12.1 m2\begin{align*} 12.1\text{ m}^2 \end{align*}

to one decimal place.

(d)

解法一

思路

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周长 ABCDAABCDA 包括 ABAB、大弧 BCDBCD 和 DADA。注意 OBOB 是半径，在图形内部，不是边界；因此 AB=AO−OB=AO−1.8AB=AO-OB=AO-1.8。

答题过程

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The major arc length BCDBCD is

1.8(2π−0.84).\begin{align*} 1.8(2\pi-0.84). \end{align*}

Also,

AB= AO−OB= 4.860…−1.8.\begin{align*} AB=&\,AO-OB\\ =&\,4.860\ldots-1.8. \end{align*}

Hence the perimeter is

AB+arc BCD+DA= (4.860…−1.8) +1.8(2π−0.84)+3.9= 16.76…\begin{align*} AB+\text{arc }BCD+DA =&\,(4.860\ldots-1.8)\\ &\,\hspace{2pt}+1.8(2\pi-0.84)+3.9\\ =&\,16.76\ldots \end{align*}

Therefore the perimeter is

16.8 m\begin{align*} 16.8\text{ m} \end{align*}

to one decimal place.