IB Mathematics SL 3.2 Use of sine, cosine and tangent ratios AI HL Paper 2- Exam Style Questions- New Syllabus

Question

The diagram shows points in a park viewed from above, at a specific moment in time. The distance between two trees, at points A and B, is \( 6.36 \, \text{m} \). Odette is playing football in the park and is standing at point O, such that \( OA = 25.9 \, \text{m} \) and \( \angle OAB = 125^\circ \).

(a) Calculate the area of triangle AOB.

Odette’s friend, Khemil, is standing at point K such that he is 12 m from A and \( \angle KAB = 45^\circ \).

(b) Calculate Khemil’s distance from B.

XY is a semicircular path in the park with centre A, such that \( \angle KAY = 45^\circ \). Khemil is standing on the path and Odette’s football is at point X. This is shown in the diagram below.

The length KX = 22.2 m , \( \angle KOX = 53.8^\circ \) and \( \angle OKX = 51.1^\circ \).

(c) Find whether Odette or Khemil is closer to the football.

(d) Khemil runs along the semicircular path to pick up the football. Calculate the distance that Khemil runs.

▶️ Answer/Explanation

Markscheme

(a)
    Attempt to use area of triangle formula (M1)
    \(\frac{1}{2} \times 25.9 \times 6.36 \times \sin(125^\circ)\)
    \( \sin(125^\circ) \approx 0.8192 \)
    \( 25.9 \times 6.36 \approx 164.724 \)
    \( 164.724 \times 0.8192 \approx 134.94 \)
    \( \frac{1}{2} \times 134.94 \approx 67.47 \)
    Note: Units are required. The final A1 is only awarded if the correct units are seen in their answer; hence award (M1)(A1)A0 for an unsupported answer of 67.5.
Result:
\( 67.5 \) m²

(b)
    Attempt to use cosine rule (M1)
    \( BK = \sqrt{12^2 + 6.36^2 – 2 \times 12 \times 6.36 \times \cos 45^\circ} \)
    \( 12^2 = 144 \), \( 6.36^2 \approx 40.4496 \)
    \( \cos 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071 \)
    \( 2 \times 12 \times 6.36 \times 0.7071 \approx 107.91 \)
    \( 144 + 40.4496 – 107.91 \approx 76.5396 \)
    \( \sqrt{76.5396} \approx 8.74738 \)
    Note: Award (M1)(A1)(A0) for radian answer of 10.2 m (10.2109\dots m) with or without working.
Result:
\( 8.75 \) m

(c)
    Method 1:
    Attempt to use sine rule with triangle OXK (M1)
    \( \frac{OX}{\sin 51.1^\circ} = \frac{22.2}{\sin 53.8^\circ} \)
    \( \sin 51.1^\circ \approx 0.7771 \), \( \sin 53.8^\circ \approx 0.8053 \)
    \( \frac{22.2}{0.8053} \approx 27.572 \)
    \( 27.572 \times 0.7771 \approx 21.4099 \)
    \( 21.4 < 22.2 \Rightarrow \) Odette is closer to the football
    Note: For final A1, 21.4 (21.4099\dots) must be seen. Follow through in part for consistent comparison.
    Method 2:

    Sketch of triangle OXK with vertices, angles, and lengths (A1)
    \( 51.1^\circ \) is the smallest angle in \( \triangle OXK \)
    Opposite side (OX) is the smallest length
    \( \therefore \) Odette is closest (R1, R1, A1)
Result:
Odette is closer to the football

(d)
    Attempt to use length of arc formula (M1)
    \( \frac{135}{360} \times 2\pi \times 12 \)
    \( \frac{135}{360} = 0.375 \)
    \( 2\pi \times 12 \approx 75.398 \)
    \( 0.375 \times 75.398 \approx 28.2743 \)
    \( = 9\pi \approx 28.3 \) m
Result:
\( 28.3 \) m

IB Mathematics SL 3.2 Use of sine, cosine and tangent ratios AI HL Paper 2- Exam Style Questions- New Syllabus

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