A pilot wishes to fly his plane from a point A to a point B on a bearing of 055°. There is a wind blowing at 120kmh⁻¹ from the west. The plane can fly at 650kmh⁻¹ in still air.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [B1] For correct triangle, may be implied by a correct sine rule or cosine rule. [diagram with vector triangle: Plane velocity (650), Wind velocity (120 from west, so East-pointing), Resultant velocity (Bearing 055)]
2. [M1] For use of a correct sine rule to obtain α = ... or θ = ... Or for a correct cosine rule leading to a value for v, followed by a correct sine rule leading to one of the other angles
3. [A1] α = 6.08° or β = 138.9 May be implied by a correct θ = awrt 49°
4. [A1] Bearing is 048.9° or 049°
5. [M1] Either v_r / sin(145 - their α) = 650 / sin(35) or v_r / sin(their α) = 120 / sin(their θ) For use of sine rule or cosine rule to find resultant velocity Do not allow for a right-angled triangle May be seen in (i)
6. [A1] v_r = 745 For correct resultant velocity, allow awrt 745
7. [M1] Time taken = 1250 / their 744.7 For correct attempt at finding time using their v, ≠ 650, 120, 770 or 530
8. [A1] = 1.68 hours or 1 hour 41 mins or 101 mins