IB Mathematics SL 3.3 Applications of right and non-right angled trigonometry AI HL Paper 2- Exam Style Questions- New Syllabus
Sara is designing a fitness trail for her school’s grounds. The diagram below displays a segment of this trail.
Sara aims to construct the trail ensuring the inner edge traces a fluid curve from point A to point B, while the outer edge follows a continuous curve from point C to point D. The distance between points A and B is 50 metres.
To form a fluid curve, Sara starts by moving to M, the midpoint of segment [AB].
(a) State the length of segment [BM]. [1]
Sara then advances 20 metres in a direction perpendicular to [AB] to arrive at point F from point M.
Point F acts as the center of a circle, with its arc defining the fluid curve linking points A and B on the trail, as illustrated in the following diagram.
(b) (i) Determine the length of segment [BF].
(ii) Calculate ∠BFM. [4]
(c) From this, determine the length of arc AB. [3]
The outer boundary of the trail, from C to D, also traces a circular arc with center F, maintaining a trail width of 2 metres.
(d) Compute the area of the curved portion of the trail, ABDC. [4]
The base of the trail will be built with concrete having a depth of 12cm.
(e) Estimate the volume of concrete needed to construct the curved portion of the trail. [3]
▶️ Answer/Explanation
(a)
Since M is the midpoint of segment [AB] and the total length of [AB] is 50 metres, the length of [BM] is half of that distance.
25 (m) [1 mark]
(b)
(i) Recognizing the need to apply the Pythagorean theorem, as [BF] forms the hypotenuse of a right triangle with legs 20 metres (perpendicular distance from M to F) and 25 metres (length of [BM]).
\[ BF^2 = 20^2 + 25^2 \]
\[ BF^2 = 400 + 625 = 1025 \]
\[ BF = \sqrt{1025} \approx 32.0156\ldots \]
Using the approximate value, \( BF = 32.0 \ (32.0156\ldots, \ \sqrt{1025}, \ 5\sqrt{41}) \ (m) \)
[2 marks]
(ii) Employing the tangent function to find \(\angle BFM\), where the opposite side is 25 metres and the adjacent side is 20 metres.
\[ \angle BFM = \tan^{-1}\!\left(\frac{25}{20}\right) \quad \text{or equivalent} \]
\[ \angle BFM = \tan^{-1}(1.25) \approx 51.3401\ldots \]
Approximating to one decimal place, \( \angle BFM = 51.3 \ (51.3401\ldots) \)
Note: Accept radian answer of 0.896 (0.896055…). Accept an answer of 51.4 from use of 3sf answer to part (b)(i) and then either cosine rule or inverse sine.
[2 marks]
Total for (b): [4 marks]
(c)
Utilizing the arc length formula, where the arc AB is part of a circle with radius BF and the central angle is twice \(\angle BFM\) (since the arc spans from A to B symmetrically around F).
\[ \text{arc length} = \frac{\theta}{360} \times 2\pi r \]
Here, \(\theta = 2 \times 51.3401\ldots = 102.6802\ldots\) degrees, and \( r = 32.0156\ldots \) metres.
\[ \text{arc length} = \frac{102.6802\ldots}{360} \times 2\pi(32.0156\ldots) \]
\[ \text{arc length} \approx \frac{102.6802}{360} \times 2 \times 3.1416 \times 32.0156 \approx 57.3755\ldots \]
Rounding to one decimal place, \( \text{arc length} = 57.4 \ (57.3755\ldots) \ (m) \)
Note: Accept 57.3 from use of 3sf values of their answers from parts (b)(i) and (b)(ii).
[3 marks]
(d)
Calculating the outer radius by adding the trail width (2 metres) to the inner radius (BF = 32.0156… metres).
Outer radius = \( 32.0156\ldots + 2 = 34.0156\ldots \) (seen anywhere)
Applying the area of a sector formula, recognizing the need to subtract the area of the inner sector from the outer sector, with the central angle of \( 102.6802\ldots \) degrees.
\[ \text{area of sector} = \frac{\theta}{360} \times \pi r^2 \]
\[ \text{area} = \frac{102.6802\ldots}{360} \times \pi \big( (34.0156\ldots)^2 – (32.0156\ldots)^2 \big) \]
\[ (34.0156\ldots)^2 – (32.0156\ldots)^2 = (34.0156\ldots – 32.0156\ldots)(34.0156\ldots + 32.0156\ldots) \]
\[ = 2 \times 66.0312\ldots \approx 132.0624\ldots \]
\[ \text{area} \approx \frac{102.6802}{360} \times 3.1416 \times 132.0624\ldots \approx 118.335\ldots \]
Rounding to the nearest whole number, \( \text{area} = 118 \ (m^2) \ (118.335\ldots) \)
[4 marks]
(e)
Multiplying the area from part (d) by the depth of the concrete (12cm = 0.12m) to find the volume.
0.12 (m) seen OR 1183350 (cm\(^2\)) seen (converting area to cm\(^2\) by multiplying by 10000).
\[ 118.335\ldots \times 0.12 \quad \text{or} \quad 1183350 \times 12 \]
\[ 118.335\ldots \times 0.12 \approx 14.2002\ldots \]
Alternatively, \( 1183350 \times 12 = 14200200 \ (14200236) \ cm^3 \), which converts to \( 14.2002\ldots \ m^3 \) (dividing by 1000000).
Rounding to one decimal place, \( 14.2 \ (14.2002\ldots) \ m^3 \quad \text{or} \quad 14200000 \ (14200236) \ cm^3 \)
[3 marks]
[Total: 15 marks]