Math  /  Trigonometry

QuestionApplication Trig Assignment 2 November 28, 2024
1. An airplane flies 210 km due south from airport A and then is diverted on a bearing of 3131^{\circ} towards airport B. Airport B is 120 km away from airport A. [ 4, 4 marks] a) On what bearing is airport B from airport A , to the nearest degree? b) How far is the airplane from airport B , to the nearest kilometre?

Studdy Solution
To find how far the airplane is from airport B, we can use the Law of Cosines in the triangle formed by the points A (airport A), C (the airplane's current position), and B (airport B).
The Law of Cosines states:
c2=a2+b22abcos(C) c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
Where: - c c is the distance from the airplane to airport B. - a=210 a = 210 km (distance from A to C). - b=120 b = 120 km (distance from A to B). - C=31 C = 31^\circ (angle at A between AC and AB).
Substitute the known values:
c2=2102+12022×210×120×cos(31) c^2 = 210^2 + 120^2 - 2 \times 210 \times 120 \times \cos(31^\circ)
Calculate c c :
c=2102+12022×210×120×cos(31) c = \sqrt{210^2 + 120^2 - 2 \times 210 \times 120 \times \cos(31^\circ)}
Using a calculator:
c44100+1440050400×cos(31) c \approx \sqrt{44100 + 14400 - 50400 \times \cos(31^\circ)} c44100+1440050400×0.8572 c \approx \sqrt{44100 + 14400 - 50400 \times 0.8572} c44100+1440043225.28 c \approx \sqrt{44100 + 14400 - 43225.28} c15274.72 c \approx \sqrt{15274.72} c123.6 c \approx 123.6
To the nearest kilometre, the airplane is approximately 124 124 km from airport B.
The bearing of airport B from airport A is 211 \boxed{211^\circ} .
The airplane is approximately 124 \boxed{124} km from airport B.

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