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

IB DP Maths AI SL Paper 2 - All Topics

Question

The diagram shows points in a public park viewed from above, at a specific moment in time.
The distance between two trees, at points A and B, is \(6.36\text{ m}\).
Sofia is kicking a soccer ball in the public park and is standing at point O, such that \( \widehat{AOB}=10^{\circ} \), \(OA=25.9\text{ m}\) and \( \widehat{OAB} \) is obtuse.
(a) Calculate the size of \( \widehat{ABO} \). [3]
(b) Calculate the area of triangle \(AOB\). [4]
Sofia’s teammate, Liam, is standing at point K such that he is \(12\text{ m}\) from A and \( \widehat{KAB}=45^{\circ} \).
(c) Calculate Liam’s distance from B. [3]
\(XY\) is a semicircular path in the park with centre A, such that \( \widehat{KAY}=45^{\circ} \). Liam is standing on the path and Sofia’s soccer ball is at point X.
The length \(KX=22.2\text{ m}\), \( \widehat{KOX}=53.8^{\circ}\) and \( \widehat{OKX}=51.1^{\circ}\).
(d) Find whether Sofia or Liam is closer to the ball. [4]
Liam runs along the semicircular path to pick up the ball.
(e) Calculate the distance that Liam runs. [3]
▶️ Answer/Explanation
Markscheme (with detailed working)

(a)

In \(\triangle AOB\), by the sine rule \[ \frac{\sin\!\big(\widehat{ABO}\big)}{\sin 10^{\circ}}=\frac{OA}{AB}=\frac{25.9}{6.36}. \] Hence \(\sin\!\big(\widehat{ABO}\big)=\sin 10^{\circ}\cdot\frac{25.9}{6.36}\approx0.7071\) so \(\boxed{\widehat{ABO}=45.0^{\circ}}\). M1 A1 A1

(b)

\(\widehat{OAB}=180^{\circ}-10^{\circ}-45^{\circ}=124.996\ldots^{\circ}\). Area \[ [AOB]=\tfrac12\,OA\cdot AB\cdot\sin\widehat{OAB} =\tfrac12(25.9)(6.36)\sin(124.996\ldots^{\circ}) \approx \boxed{67.5\ \text{m}^2}. \] A1 M1 A1 (units) A1

(c)

In \(\triangle ABK\) with \(AK=12\), \(AB=6.36\), included angle \(45^{\circ}\), \[ BK^2=12^2+6.36^2-2(12)(6.36)\cos45^{\circ} \Rightarrow BK\approx \boxed{8.75\ \text{m}}. \] M1 A1 A1

(d)

In \(\triangle OKX\), by the sine rule \[ \frac{OX}{\sin 53.8^{\circ}}=\frac{KX}{\sin 51.1^{\circ}} \Rightarrow OX=\frac{\sin 53.8^{\circ}}{\sin 51.1^{\circ}}\cdot 22.2 \approx 21.41\ \text{m}. \] Since \(OX\approx 21.4\text{ m} < KX=22.2\text{ m}\), \(\boxed{\text{Sofia is closer to the ball}}\). M1 A1 A1 A1

(e)

The arc from \(K\) to \(X\) subtends \(135^{\circ}\) at centre \(A\) on a circle of radius \(AK=12\). Arc length \(=\dfrac{135}{360}\cdot 2\pi(12)=\boxed{9\pi\ \text{m}\ (\approx 28.3\ \text{m})}\). M1 A1 A1
Total Marks: 17
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