Zadanie 2e, Paper 3, maj 2025, TZ2, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2025, TZ2, Paper 3 (zadania wzorowane), pytanie 2(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Three fictional research stations are called Westhaven (W), Meridian (M), and Relay (N). Compare a flat planning map with a spherical model. Bearings are measured clockwise from north.
Polecenie
In the spherical model the world has radius 5000 km. Coordinates are in thousands of kilometres, with origin at its centre. The north pole P and station N have vectors and . The positive x direction points west along the equator from N. An equatorial reference point E has vector . Station M lies north of N on the same meridian, with and 0 < θ < π/2. The shorter surface distance NM is 3500 km. Station W is on the equator west of N, with . Surface distances follow the shorter great-circle arc.
Find the shortest surface distance between W and M.
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Odpowiedź z klucza
Let α be their central angle. The dot-product formula gives . Thus α = 1.770068260… rad, or 101.4174408…°. The shortest arc has length 5000α = 8850.341302… km, hence 8850 km. A cross-product sine method must select the obtuse central angle because the dot product is negative.
Schemat punktowania
M1: use scalar product, chord/cosine rule or a cross-product formula for the central angle. A1: correct substituted relation. A1: the obtuse central angle. M1: convert the angle to arc length. A1: 8850 km. If an inverse-sine method retains only its acute principal value, the two angle/distance accuracy marks are unavailable; the other three marks may be earned.
Komentarz. Paper 3: calculators permitted. Give reproducible algebra or numerical work. Round final numerical answers to three significant figures unless stated otherwise.